blocks|key|821433|text|Dyalog+Unicode，10字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|821434|1⊥⍳×⍳×1-⍨⍳|code-block|syntax|javascript|821435|在网上试试！|unstyled|821436|是如何工作的|821437|1⊥⍳×⍳×1-⍨⍳
++⍳×⍳×1-⍨⍳++Compute+(x%5E3+-+x%5E2)+for+1..n
1⊥++++++++++Sum|821438|使用“和的平方”等于“立方和”的事实。|entityMap|0|LINK|mutability|MUTABLE|url|https://www.dyalog.com/|1|https://tio.run/##SyzI0U2pTMzJT///3/BR19JHvZsPTwcThrqPelcAWf/THrVNeNTb96hvqqf/o67mQ@uNH7VNBPKCg5yBZIiHZ/D/NAUjrjQFYyA2NAAA^0|0|E|0|0|0|0|6|1|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|L|3|M|5|6|7|19|8|@]|9|@]|C|$]]|$1|N|3|O|5|F|7|1A|8|@]|9|@]|C|$G|H]]|$1|P|3|Q|5|K|7|1B|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|T|U|V|C|$W|Z]]]]

<h1><a href="https://www.dyalog.com/" rel="noreferrer">APL (Dyalog Unicode)</a>, 10 bytes</h1>



<pre class="lang-apl prettyprint-override"><code>1⊥⍳×⍳×1-⍨⍳
</code></pre>

<p><a href="https://tio.run/##SyzI0U2pTMzJT///3/BR19JHvZsPTwcThrqPelcAWf/THrVNeNTb96hvqqf/o67mQ@uNH7VNBPKCg5yBZIiHZ/D/NAUjrjQFYyA2NAAA" rel="noreferrer" title="APL (Dyalog Unicode) – Try It Online">Try it online!</a></p>

<h3>How it works</h3>

<pre class="lang-apl prettyprint-override"><code>1⊥⍳×⍳×1-⍨⍳
  ⍳×⍳×1-⍨⍳  Compute (x^3 - x^2) for 1..n
1⊥          Sum
</code></pre>

<p>Uses the fact that "square of sum" is equal to "sum of cubes".</p>


blocks|key|821262|text|JavaScript，20字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|821263|f=n=>n&&n*n*--n%2Bf(n)|code-block|syntax|javascript|821264|在网上试试|unstyled|offset|length|entityMap|0|LINK|mutability|MUTABLE|url|https://tio.run/##y0osSyxOLsosKNHNy09J/f8/zTbP1i5PTS1PK09LVzdPO00jT/N/cn5ecX5Oql5OfrpGmoaRiabmfwA^0|0|0|0|5|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|U|8|@]|9|@$J|V|K|W|1|X]]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<h1>JavaScript, 20 bytes</h1>
<pre><code>f=n=&gt;n&amp;&amp;n*n*--n+f(n)
</code></pre>
<p><a href="https://tio.run/##y0osSyxOLsosKNHNy09J/f8/zTbP1i5PTS1PK09LVzdPO00jT/N/cn5ecX5Oql5OfrpGmoaRiabmfwA" rel="noreferrer">Try it online</a></p>


blocks|key|821574|text|数据中心，16字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|821575|?dd3%5Er-r3*2%2B*C/p|code-block|syntax|javascript|821576|实现(n%5E3-n)(3n%2B2)/12|unstyled|style|CODE|821577|在网上试试！|entityMap|0|LINK|mutability|MUTABLE|url|https://www.gnu.org/software/bc/manual/dc-1.05/html_mono/dc.html|1|INLINETEX|IMMUTABLE|teX|(n%5E3-n)(3n%2B2)/12|2|https://tio.run/##S0n@/98@JcU4rki3yFjLSFvLWb/g/39DAwA^0|0|4|0|0|0|2|G|2|G|1|0|0|6|2^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@$A|14|B|15|1|16]]|C|$]]|$1|D|3|E|5|F|7|17|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|18|8|@$A|19|B|1A|L|M]]|9|@$A|1B|B|1C|1|1D]]|C|$]]|$1|N|3|O|5|K|7|1E|8|@]|9|@$A|1F|B|1G|1|1H]]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|X|S|Y|C|$Z|10]]|11|$5|R|S|T|C|$U|12]]]]

<h1><a href="https://www.gnu.org/software/bc/manual/dc-1.05/html_mono/dc.html" rel="nofollow noreferrer">dc</a>, 16 bytes</h1>

<pre><code>?dd3^r-r3*2+*C/p
</code></pre>

<p>Implements <span class="math-container">\$(n^3-n)(3n+2)/12\$</span></p>

<p><a href="https://tio.run/##S0n@/98@JcU4rki3yFjLSFvLWb/g/39DAwA" rel="nofollow noreferrer" title="dc – Try It Online">Try it online!</a></p>


blocks|key|817035|text|APL(Dyalog)，17字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|817036|{%2B/(¯1↓⍵)×1↓×⍨⍵}⍳|code-block|syntax|javascript|817037|乔纳森·艾伦港的果冻回答。|unstyled|817038|在网上试试！|offset|length|entityMap|0|LINK|mutability|MUTABLE|url|https://tio.run/##SyzI0U2pTMzJT///v1pbX@PQesNHbZMf9W7VPDwdxDo8/VHvCiC39lHv5v9pj9omPOrte9Q31dP/UVfzofXGj9omAnnBQc5AMsTDM/h/moIRV5qCMRAbGgAA^0|0|0|0|0|6|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|W|8|@]|9|@]|A|$]]|$1|J|3|K|5|I|7|X|8|@]|9|@$L|Y|M|Z|1|10]]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

<h1>APL(Dyalog), 17 bytes</h1>

<pre><code>{+/(¯1↓⍵)×1↓×⍨⍵}⍳
</code></pre>

<p>(Much longer) Port of Jonathan Allan's Jelly answer.</p>

<p><a href="https://tio.run/##SyzI0U2pTMzJT///v1pbX@PQesNHbZMf9W7VPDwdxDo8/VHvCiC39lHv5v9pj9omPOrte9Q31dP/UVfzofXGj9omAnnBQc5AMsTDM/h/moIRV5qCMRAbGgAA" rel="nofollow noreferrer">Try it online!</a></p>


blocks|key|817088|text|[化]双模拟，16字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|817089|((×⍨%2B/)-(%2B/×⍨))⍳|code-block|syntax|javascript|817090|在网上试试！|unstyled|817091|+(×⍨%2B/)++++++++++++The+square+(×+self)+of+the+sum+(%2B+fold)
+++++++-+++++++++++minus
++++++++(%2B/×⍨)+++++the+sum+of+the+square
(+++++++++++++)⍳+++of+[1,+2,+…+input].|entityMap|0|LINK|mutability|MUTABLE|url|https://www.dyalog.com/|1|https://tio.run/##SyzI0U2pTMzJT///X0Pj8PRHvSu09TV1NbT1wWxNzUe9m/@nPWqb8Ki371HfVE//R13Nh9YbP2qbCOQFBzkDyRAPz@D/aQpGXGkKxkBsaAAA^0|0|6|0|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|L|3|M|5|F|7|15|8|@]|9|@]|C|$G|H]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

<h1><a href="https://www.dyalog.com/" rel="nofollow noreferrer">APL (Dyalog)</a>, 16 bytes</h1>



<pre class="lang-apl prettyprint-override"><code>((×⍨+/)-(+/×⍨))⍳
</code></pre>

<p><a href="https://tio.run/##SyzI0U2pTMzJT///X0Pj8PRHvSu09TV1NbT1wWxNzUe9m/@nPWqb8Ki371HfVE//R13Nh9YbP2qbCOQFBzkDyRAPz@D/aQpGXGkKxkBsaAAA" rel="nofollow noreferrer" title="APL (Dyalog Unicode) – Try It Online">Try it online!</a></p>

<pre class="lang-apl prettyprint-override"><code> (×⍨+/)            The square (× self) of the sum (+ fold)
       -           minus
        (+/×⍨)     the sum of the square
(             )⍳   of [1, 2, … input].
</code></pre>


blocks|key|817316|text|Java+(JDK)，23字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|817317|n->(3*n%2B2)*(n*n*n-n)/12|code-block|syntax|javascript|817318|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|http://jdk.java.net/|1|https://tio.run/##NY5Na8MwDIbv@RWiELAzx2vSnlo62HGHsUPZaezgJXFxlsjGVgqh5LdnTsIQQl/PK6lVd5W39e9semc9QRtrOZDppB6wImNRZuek6lQI8K4MwiMBcMNPZyoIpCiGuzU19HHGruQN3r6@Qflb4CsK8Ib0icqPH67xiqwHfZkxf2GHDJ9KnjHMouXIn4tyPq8KgxR3UBMowAUehYBSwEFAsY/ZcYn7aSO19SzSK3vaFP9nAa5joKaXdiDp4l@k2S491idI6xR3YoEFaKmc68bXEL9kS4vzbfOULD7Nfw^0|0|A|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<h1><a href="http://jdk.java.net/" rel="nofollow noreferrer">Java (JDK)</a>, 23 bytes</h1>



<pre class="lang-java prettyprint-override"><code>n-&gt;(3*n+2)*(n*n*n-n)/12
</code></pre>

<p><a href="https://tio.run/##NY5Na8MwDIbv@RWiELAzx2vSnlo62HGHsUPZaezgJXFxlsjGVgqh5LdnTsIQQl/PK6lVd5W39e9semc9QRtrOZDppB6wImNRZuek6lQI8K4MwiMBcMNPZyoIpCiGuzU19HHGruQN3r6@Qflb4CsK8Ib0icqPH67xiqwHfZkxf2GHDJ9KnjHMouXIn4tyPq8KgxR3UBMowAUehYBSwEFAsY/ZcYn7aSO19SzSK3vaFP9nAa5joKaXdiDp4l@k2S491idI6xR3YoEFaKmc68bXEL9kS4vzbfOULD7Nfw" rel="nofollow noreferrer" title="Java (JDK) – Try It Online">Try it online!</a></p>


blocks|key|817350|text|05AB1E，8字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|817351|ÝDOnsnO-|code-block|syntax|javascript|817352|解释：|unstyled|817353|ÝDOnsnO-+++++//Full+program
Ý++++++++++++//Push+[0..a]+where+a+is+implicit+input
+D+++++++++++//Duplicate+top+of+stack
++On+++++++++//Push+sum,+then+square+it
++++s++++++++//Swap+top+two+elements+of+stack
+++++nO++++++//Square+each+element,+then+push+sum
+++++++-+++++//Difference+(implicitly+printed)|817354|在网上试试！|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Adriandmen/05AB1E|1|https://tio.run/##yy9OTMpM/f//8FwX/7ziPH/d//8NDQA^0|0|6|0|0|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|13|8|@]|9|@]|C|$]]|$1|L|3|M|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|N|3|O|5|K|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

<h1><a href="https://github.com/Adriandmen/05AB1E" rel="nofollow noreferrer">05AB1E</a>, 8 bytes</h1>

<pre><code>ÝDOnsnO-
</code></pre>

<p>Explanation:</p>

<pre><code>ÝDOnsnO-     //Full program
Ý            //Push [0..a] where a is implicit input
 D           //Duplicate top of stack
  On         //Push sum, then square it
    s        //Swap top two elements of stack
     nO      //Square each element, then push sum
       -     //Difference (implicitly printed)
</code></pre>

<p><a href="https://tio.run/##yy9OTMpM/f//8FwX/7ziPH/d//8NDQA" rel="nofollow noreferrer" title="05AB1E – Try It Online">Try it online!</a></p>


blocks|key|822049|text|MathGolf，6字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|822050|{î²ï*%2B|code-block|syntax|javascript|822051|在网上试试！|unstyled|822052|计算\sum_{k=1}%5En+(k%5E2(k-1))|style|CODE|822053|解释：|822054|{+++++++Loop+(implicit)+input+times
+î²+++++1-index+of+loop+squared
++++*+++Multiplied+by
+++ï++++The+0-index+of+the+loop
+++++%2B++And+add+to+the+running+total|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/maxbergmark/mathgolf|1|https://tio.run/##y00syUjPz0n7/7/68LpDmw6v19L@/9@Qy4jLmMvQgMvIBEgaAAA|2|INLINETEX|IMMUTABLE|teX|\sum_{k=1}%5En+(k%5E2(k-1))^0|0|8|0|0|0|0|6|1|0|2|N|2|N|2|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@$A|18|B|19|1|1A]]|C|$]]|$1|D|3|E|5|F|7|1B|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1C|8|@]|9|@$A|1D|B|1E|1|1F]]|C|$]]|$1|L|3|M|5|K|7|1G|8|@$A|1H|B|1I|N|O]]|9|@$A|1J|B|1K|1|1L]]|C|$]]|$1|P|3|Q|5|6|7|1M|8|@]|9|@]|C|$]]|$1|R|3|S|5|F|7|1N|8|@]|9|@]|C|$G|H]]]|T|$U|$5|V|W|X|C|$Y|Z]]|10|$5|V|W|X|C|$Y|11]]|12|$5|13|W|14|C|$15|16]]]]

<h1><a href="https://github.com/maxbergmark/mathgolf" rel="nofollow noreferrer">MathGolf</a>, 6 bytes</h1>

<pre><code>{î²ï*+
</code></pre>

<p><a href="https://tio.run/##y00syUjPz0n7/7/68LpDmw6v19L@/9@Qy4jLmMvQgMvIBEgaAAA" rel="nofollow noreferrer" title="MathGolf – Try It Online">Try it online!</a></p>

<p>Calculates <span class="math-container">\$\sum_{k=1}^n (k^2(k-1))\$</span></p>

<h3>Explanation:</h3>

<pre><code>{       Loop (implicit) input times
 î²     1-index of loop squared
    *   Multiplied by
   ï    The 0-index of the loop
     +  And add to the running total
</code></pre>


blocks|key|821238|text|JavaScript+(ES6)，22字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|821239|n=>n*~-n*-~n*(n/4%2B1/6)|code-block|syntax|javascript|821240|在网上试试！|unstyled|offset|length|entityMap|0|LINK|mutability|MUTABLE|url|https://tio.run/##y0osSyxOLsosKNHNy09J/Z9m@z/P1i5Pq043T0u3Lk9LI0/fRNtQ30zzf3J@XnF@TqpeTn66RpqGoaYmF6qIEYaIMYaIoQGmNhMsqoDK/gMA^0|0|0|0|6|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|U|8|@]|9|@$J|V|K|W|1|X]]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<h1>JavaScript (ES6), 22 bytes</h1>

<pre class="lang-javascript prettyprint-override"><code>n=&gt;n*~-n*-~n*(n/4+1/6)
</code></pre>
<p><a href="https://tio.run/##y0osSyxOLsosKNHNy09J/Z9m@z/P1i5Pq043T0u3Lk9LI0/fRNtQ30zzf3J@XnF@TqpeTn66RpqGoaYmF6qIEYaIMYaIoQGmNhMsqoDK/gMA" rel="nofollow noreferrer" title="JavaScript (Node.js) – Try It Online">Try it online!</a></p>


blocks|key|817123|text|Pyth，7字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|817124|sm**hdh|code-block|syntax|javascript|817125|试一试在线这里。|unstyled|offset|length|817126|使用尼尔的回答中的公式。|817127|sm**hdhddQ+++Implicit:+Q=eval(input())
+++++++++++++Trailing+ddQ+inferred
+m+++++++Q+++Map+[0-Q)+as+d,+using:
++++hd+++++++++Increment+d
+++*++hd+++++++Multiply+the+above+with+another+copy
++*+++++d++++++Multiply+the+above+by+d
s++++++++++++Sum,+implicit+print+|entityMap|0|LINK|mutability|MUTABLE|url|https://pyth.herokuapp.com/?code=sm%252A%252Ahdh&input=10&debug=0|1|https://codegolf.stackexchange.com/a/175268/41020^0|0|0|5|2|0|0|2|5|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|10|8|@]|9|@$J|11|K|12|1|13]]|A|$]]|$1|L|3|M|5|I|7|14|8|@]|9|@$J|15|K|16|1|17]]|A|$]]|$1|N|3|O|5|D|7|18|8|@]|9|@]|A|$E|F]]]|P|$Q|$5|R|S|T|A|$U|V]]|W|$5|R|S|T|A|$U|X]]]]

<h1>Pyth, 7 bytes</h1>
<pre><code>sm**hdh
</code></pre>
<p>Try it online <a href="https://pyth.herokuapp.com/?code=sm%2A%2Ahdh&amp;input=10&amp;debug=0" rel="nofollow noreferrer">here</a>.</p>
<p>Uses the formula in <a href="https://codegolf.stackexchange.com/a/175268/41020">Neil's answer</a>.</p>
<pre><code>sm**hdhddQ   Implicit: Q=eval(input())
             Trailing ddQ inferred
 m       Q   Map [0-Q) as d, using:
    hd         Increment d
   *  hd       Multiply the above with another copy
  *     d      Multiply the above by d
s            Sum, implicit print 
</code></pre>


blocks|key|821460|text|帕里/GP，21字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|821461|n->(3*n%2B2)*(n%5E3-n)/12|code-block|syntax|javascript|821462|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|http://pari.math.u-bordeaux.fr/|1|https://tio.run/##FcdLCoAwDEXRrTw6ajTFfpzqRkTBSUWQEIoTVx91dM/VvZ3hUKuYYBJmXzrpM3VethKEhpRtV70eLwgztJ1yf3T/OFQvRIwlMTKjMFL8NP6NK9kL^0|0|5|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<h1><a href="http://pari.math.u-bordeaux.fr/" rel="nofollow noreferrer">Pari/GP</a>, 21 bytes</h1>

<pre><code>n-&gt;(3*n+2)*(n^3-n)/12
</code></pre>

<p><a href="https://tio.run/##FcdLCoAwDEXRrTw6ajTFfpzqRkTBSUWQEIoTVx91dM/VvZ3hUKuYYBJmXzrpM3VethKEhpRtV70eLwgztJ1yf3T/OFQvRIwlMTKjMFL8NP6NK9kL" rel="nofollow noreferrer" title="Pari/GP – Try It Online">Try it online!</a></p>


blocks|key|817630|text|05AB1E，6字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|817631|LnDƶαO|code-block|syntax|javascript|817632|在网上试试！|unstyled|817633|解释|817634|L+++++++++#+push+range+[1+...+input]
+n++++++++#+square+each
++D+++++++#+duplicate
+++ƶ++++++#+lift,+multiply+each+by+its+1-based+index
++++α+++++#+element-wise+absolute+difference
+++++O++++#+sum|817635|在相同字节数下的其他一些版本：|817636|L<ān*O|style|CODE|817637|817638|Ln.āPO|817639|817640|L¦nā*O|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Adriandmen/05AB1E|1|https://tio.run/##yy9OTMpM/f/fJ8/l2LZzG/3//zc0MAAA^0|0|6|0|0|0|0|6|1|0|0|0|0|0|6|0|0|0|6|0|0|0|6^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@$A|1B|B|1C|1|1D]]|C|$]]|$1|D|3|E|5|F|7|1E|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1F|8|@]|9|@$A|1G|B|1H|1|1I]]|C|$]]|$1|L|3|M|5|6|7|1J|8|@]|9|@]|C|$]]|$1|N|3|O|5|F|7|1K|8|@]|9|@]|C|$G|H]]|$1|P|3|Q|5|K|7|1L|8|@]|9|@]|C|$]]|$1|R|3|S|5|K|7|1M|8|@$A|1N|B|1O|T|U]]|9|@]|C|$]]|$1|V|3|-4|5|K|7|1P|8|@]|9|@]|C|$]]|$1|W|3|X|5|K|7|1Q|8|@$A|1R|B|1S|T|U]]|9|@]|C|$]]|$1|Y|3|-4|5|K|7|1T|8|@]|9|@]|C|$]]|$1|Z|3|10|5|K|7|1U|8|@$A|1V|B|1W|T|U]]|9|@]|C|$]]]|11|$12|$5|13|14|15|C|$16|17]]|18|$5|13|14|15|C|$16|19]]]]

<h1><a href="https://github.com/Adriandmen/05AB1E" rel="nofollow noreferrer">05AB1E</a>, 6 bytes</h1>

<pre><code>LnDƶαO
</code></pre>

<p><a href="https://tio.run/##yy9OTMpM/f/fJ8/l2LZzG/3//zc0MAAA" rel="nofollow noreferrer" title="05AB1E – Try It Online">Try it online!</a></p>

<p><strong>Explanation</strong></p>

<pre><code>L         # push range [1 ... input]
 n        # square each
  D       # duplicate
   ƶ      # lift, multiply each by its 1-based index
    α     # element-wise absolute difference
     O    # sum
</code></pre>

<p>Some other versions at the same byte count:</p>

<p><code>L&lt;ān*O</code><br>
<code>Ln.āPO</code><br>
<code>L¦nā*O</code>  </p>


blocks|key|821664|text|R，28字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|821665|x=1:scan();sum(x)%5E2-sum(x%5E2)|code-block|syntax|javascript|821666|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://www.r-project.org/|1|https://tio.run/##K/r/v8LW0Ko4OTFPQ9O6uDRXo0IzzkgXzIgz0vxvaGDw/z8A^0|0|1|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<h1><a href="https://www.r-project.org/" rel="nofollow noreferrer">R</a>, 28 bytes</h1>



<pre class="lang-r prettyprint-override"><code>x=1:scan();sum(x)^2-sum(x^2)
</code></pre>

<p><a href="https://tio.run/##K/r/v8LW0Ko4OTFPQ9O6uDRXo0IzzkgXzIgz0vxvaGDw/z8A" rel="nofollow noreferrer" title="R – Try It Online">Try it online!</a></p>


blocks|key|822863|text|MATLAB/Octave，21字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|822864|@(x)(x%5E3-x)*(x/4%2B1/6)|code-block|syntax|javascript|822865|匿名函数。如果我们假设x可以作为工作空间中的变量给出，那么我们可以另外保存4个字节。|unstyled|offset|length|style|CODE|822866|822867|实现n(n−1)(n%2B1)(3n%2B2)/12|822868|我的方法之一是@(x)-sum([sum(1:x)*i,1:x].%5E2)，它长了几个字节，但是我发现它很有趣，因为它使用假想的单位来改变符号。|822869|822870|我还发现，我可以通过与(1:x)*(1:x)'相乘法向量和转置向量来生成平方和。可能对未来的挑战有用！|entityMap|0|INLINETEX|mutability|IMMUTABLE|teX|n(n−1)(n%2B1)(3n%2B2)/12^0|0|0|B|1|0|0|2|K|2|K|0|0|7|T|0|0|B|C^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|14|8|@$J|15|K|16|L|M]]|9|@]|A|$]]|$1|N|3|-4|5|I|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|I|7|18|8|@$J|19|K|1A|L|M]]|9|@$J|1B|K|1C|1|1D]]|A|$]]|$1|Q|3|R|5|I|7|1E|8|@$J|1F|K|1G|L|M]]|9|@]|A|$]]|$1|S|3|-4|5|I|7|1H|8|@]|9|@]|A|$]]|$1|T|3|U|5|I|7|1I|8|@$J|1J|K|1K|L|M]]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

<h1>MATLAB/Octave, 21 bytes</h1>
<pre><code>@(x)(x^3-x)*(x/4+1/6)
</code></pre>
<p>Anonymous function. We could save another 4 bytes if we assume <code>x</code> can be given as variable in workspace.<br />
Implements <span class="math-container">\$n(n−1)(n+1)(3n+2)/12\$</span></p>
<hr />
<p>One of my approaches was <code>@(x)-sum([sum(1:x)*i,1:x].^2)</code>
which is few bytes longer but I found it interesting because It used imaginary unit to change the sign.<br />
I also found out I can generate sum of squares by multiplying normal and transposed vector with <code>(1:x)*(1:x)'</code>. Might be useful for future challenges!</p>


blocks|key|819136|text|迷宫，20字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|819137|?(:):):_3*(***_12/!@|code-block|syntax|javascript|819138|在网上试试！|unstyled|819139|(x-1)x(x%2B1)(3x%2B2)/12的直截了当的闭形公式。|style|CODE|819140|?++++++Take+input+[x]
(:):)++Decrement,+dup,+increment,+dup,+increment+[x-1+x+x%2B1]
:_3*(++Dup,+3+times,+decrement+[x-1+x+x%2B1+3x%2B2]
***++++Multiply+four+values
_12/+++Divide+by+12
!@+++++Print+it+and+halt|819141|迷宫，21字节|819142|?
::(";;;!@
{++:
%2B**}|819143|819144|迷宫，22字节|819145|+++?
:(::
}++"
**%2B{;!@|819146|819147|两个版本使用一个循环来计算\sum_{i=1}%5E{x}{i%5E2(i-1)}。这两个版本都用i-1+==+0测试循环退出，必要时丢弃顶部值;，打印和退出。|819148|?+++++Take+input+[x];+enter+loop+[sum+i]+with+initial+values+of+sum=0,+i=x
::(+++Dup,+dup,+decrement+[sum+i+i+i-1]
:}++++Dup+and+move+i-1+to+aux.+store+[sum+i+i+i-1+%7C+i-1]
**%2B+++Multiply+top+three+values+and+add+to+sum+[sum'+%7C+i-1]
{+++++Move+i-1+back+to+the+main+stack

++++++Loop+exit+test+can+be+done+anywhere+in+the+last+iteration
++++++(i==1+or+i==0),+since+that+iteration+does+not+affect+the+sum.|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/m-ender/labyrinth|1|https://tio.run/##y0lMqizKzCvJ@P/fXsNKEwjjjbU0tLS04g2N9BUd/v83NDAAAA|2|INLINETEX|IMMUTABLE|teX|(x-1)x(x%2B1)(3x%2B2)/12|3|4|https://tio.run/##y0lMqizKzCvJ@P/fnsvKSkPJ2tpa0YGrWkHBiktbS6v2/39DAwMA|5|6|https://tio.run/##y0lMqizKzCvJ@P9fQUHBnstKw8qKq1ZBQYlLS0u72lrR4f9/QwMDAA|7|\sum_{i=1}%5E{x}{i%5E2(i-1)}^0|0|2|0|0|0|0|6|1|0|0|K|0|K|2|0|0|0|2|3|0|0|0|6|4|0|0|2|5|0|0|0|6|6|0|D|O|19|8|1W|1|D|O|7|0^^$0|@$1|2|3|4|5|6|7|1R|8|@]|9|@$A|1S|B|1T|1|1U]]|C|$]]|$1|D|3|E|5|F|7|1V|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1W|8|@]|9|@$A|1X|B|1Y|1|1Z]]|C|$]]|$1|L|3|M|5|K|7|20|8|@$A|21|B|22|N|O]]|9|@$A|23|B|24|1|25]]|C|$]]|$1|P|3|Q|5|F|7|26|8|@]|9|@]|C|$G|H]]|$1|R|3|S|5|6|7|27|8|@]|9|@$A|28|B|29|1|2A]]|C|$]]|$1|T|3|U|5|F|7|2B|8|@]|9|@]|C|$G|H]]|$1|V|3|J|5|K|7|2C|8|@]|9|@$A|2D|B|2E|1|2F]]|C|$]]|$1|W|3|X|5|6|7|2G|8|@]|9|@$A|2H|B|2I|1|2J]]|C|$]]|$1|Y|3|Z|5|F|7|2K|8|@]|9|@]|C|$G|H]]|$1|10|3|J|5|K|7|2L|8|@]|9|@$A|2M|B|2N|1|2O]]|C|$]]|$1|11|3|12|5|K|7|2P|8|@$A|2Q|B|2R|N|O]|$A|2S|B|2T|N|O]|$A|2U|B|2V|N|O]]|9|@$A|2W|B|2X|1|2Y]]|C|$]]|$1|13|3|14|5|F|7|2Z|8|@]|9|@]|C|$G|H]]]|15|$16|$5|17|18|19|C|$1A|1B]]|1C|$5|17|18|19|C|$1A|1D]]|1E|$5|1F|18|1G|C|$1H|1I]]|1J|$5|17|18|19|C|$1A|1B]]|1K|$5|17|18|19|C|$1A|1L]]|1M|$5|17|18|19|C|$1A|1B]]|1N|$5|17|18|19|C|$1A|1O]]|1P|$5|1F|18|1G|C|$1H|1Q]]]]

<h1><a href="https://github.com/m-ender/labyrinth" rel="nofollow noreferrer">Labyrinth</a>, 20 bytes</h1>
<pre><code>?(:):):_3*(***_12/!@
</code></pre>
<p><a href="https://tio.run/##y0lMqizKzCvJ@P/fXsNKEwjjjbU0tLS04g2N9BUd/v83NDAAAA" rel="nofollow noreferrer" title="Labyrinth – Try It Online">Try it online!</a></p>
<p>Straightforward closed-form formula of <span class="math-container">\$(x-1)x(x+1)(3x+2)/12\$</span>.</p>
<pre><code>?      Take input [x]
(:):)  Decrement, dup, increment, dup, increment [x-1 x x+1]
:_3*(  Dup, 3 times, decrement [x-1 x x+1 3x+2]
***    Multiply four values
_12/   Divide by 12
!@     Print it and halt
</code></pre>
<h1><a href="https://github.com/m-ender/labyrinth" rel="nofollow noreferrer">Labyrinth</a>, 21 bytes</h1>
<pre><code>?
::(&quot;;;;!@
{  :
+**}
</code></pre>
<p><a href="https://tio.run/##y0lMqizKzCvJ@P/fnsvKSkPJ2tpa0YGrWkHBiktbS6v2/39DAwMA" rel="nofollow noreferrer" title="Labyrinth – Try It Online">Try it online!</a></p>
<h1><a href="https://github.com/m-ender/labyrinth" rel="nofollow noreferrer">Labyrinth</a>, 22 bytes</h1>
<pre><code>   ?
:(::
}  &quot;
**+{;!@
</code></pre>
<p><a href="https://tio.run/##y0lMqizKzCvJ@P9fQUHBnstKw8qKq1ZBQYlLS0u72lrR4f9/QwMDAA" rel="nofollow noreferrer" title="Labyrinth – Try It Online">Try it online!</a></p>
<p>Two versions using a loop to evaluate <span class="math-container">\$\sum_{i=1}^{x}{i^2(i-1)}\$</span>. Both versions test the loop exit with <code>i-1 == 0</code>, discard the top values <code>;</code> as necessary, print the sum and exit.</p>
<pre><code>?     Take input [x]; enter loop [sum i] with initial values of sum=0, i=x
::(   Dup, dup, decrement [sum i i i-1]
:}    Dup and move i-1 to aux. store [sum i i i-1 | i-1]
**+   Multiply top three values and add to sum [sum' | i-1]
{     Move i-1 back to the main stack

      Loop exit test can be done anywhere in the last iteration
      (i==1 or i==0), since that iteration does not affect the sum.
</code></pre>


blocks|key|819246|text|ARM+(拇指)，18字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|819247|.section+.text
.global+func
.thumb
//+int+func(int+dummy,+int+x)
//+returns+sum(i*i*(i-1)+for+i+in+1..x)
//+r1+=+i+(x->1),+r2+=+i*i*(i-1),+r0+=+sum
func:
++++mov+r0,+#0+++++++//+2000+++int+sum+=+0;
++++mov+r2,+#0+++++++//+2200+++int+prod+=+0;
.loop:+++++++++++++++//++++++++do+{
++++add+r0,+r0,+r2+++//+1880+++sum+%2B=+prod;
++++mov+r2,+r1+++++++//+1c0a+++prod+=+x;
++++mul+r2,+r1+++++++//+434a+++prod+*=+x;
++++sub+r1,+#1+++++++//+3901+++x+-=+1;
++++mul+r2,+r1+++++++//+434a+++prod+*=+x;
++++bne+.loop++++++++//+d1f9+++}+while+(prod+!=+0);
++++bx+lr++++++++++++//+4770+++return+sum;|code-block|syntax|javascript|819248|一个通过r1获取输入整数并通过r0返回值的函数。|unstyled|offset|length|style|CODE|819249|使用迭代公式\sum_{i=1}%5E{x}{i%5E2(i-1)}，向后迭代。当i%5E2(i-1)的值为0时，循环终止。|819250|为了编写测试用例，C端函数签名采用r0的虚拟值。|819251|用C编写的测试用例：|819252|#include+<stdio.h>

int+func(int+dummy,+int+x);
int+main(void)+{
++++int+dummy;
++++int+inputs[]+=+{1,+2,+3,+10,+24,+100};
++++for(int+i+=+0;+i+<+6;+%2B%2Bi)+{
++++++++int+v+=+inputs[i];
++++++++printf("input+=+%25d+->+ans+=+%25d\n",+v,+func(dummy,+v));
++++}
++++return+0;
}|819253|生成、运行、objdump命令：|819254|arm-linux-gnueabihf-gcc+main_c.c+func_c.S+-static+-o+main_c
qemu-arm+-L+/usr/arm-linux-gnueabihf+./main_c
arm-linux-gnueabihf-objdump+-d+main_c+%7C+awk+-v+RS=+'/%5E[[:xdigit:]]%2B+<(func%7C.loop)>/'|819255|(+objdump+%2B+awk技巧来自这就是答案。)|entityMap|0|INLINETEX|mutability|IMMUTABLE|teX|\sum_{i=1}%5E{x}{i%5E2(i-1)}|1|i%5E2(i-1)|2|LINK|MUTABLE|url|https://stackoverflow.com/a/31138400^0|0|0|4|2|F|2|0|6|O|11|8|6|O|0|11|8|1|0|H|2|0|0|0|0|0|J|5|2^^$0|@$1|2|3|4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|1H|8|@$J|1I|K|1J|L|M]|$J|1K|K|1L|L|M]]|9|@]|A|$]]|$1|N|3|O|5|I|7|1M|8|@$J|1N|K|1O|L|M]|$J|1P|K|1Q|L|M]]|9|@$J|1R|K|1S|1|1T]|$J|1U|K|1V|1|1W]]|A|$]]|$1|P|3|Q|5|I|7|1X|8|@$J|1Y|K|1Z|L|M]]|9|@]|A|$]]|$1|R|3|S|5|I|7|20|8|@]|9|@]|A|$]]|$1|T|3|U|5|D|7|21|8|@]|9|@]|A|$E|F]]|$1|V|3|W|5|I|7|22|8|@]|9|@]|A|$]]|$1|X|3|Y|5|D|7|23|8|@]|9|@]|A|$E|F]]|$1|Z|3|10|5|I|7|24|8|@]|9|@$J|25|K|26|1|27]]|A|$]]]|11|$12|$5|13|14|15|A|$16|17]]|18|$5|13|14|15|A|$16|19]]|1A|$5|1B|14|1C|A|$1D|1E]]]]

<h1>ARM (Thumb), 18 bytes</h1>
<pre><code>.section .text
.global func
.thumb
// int func(int dummy, int x)
// returns sum(i*i*(i-1) for i in 1..x)
// r1 = i (x-&gt;1), r2 = i*i*(i-1), r0 = sum
func:
    mov r0, #0       // 2000   int sum = 0;
    mov r2, #0       // 2200   int prod = 0;
.loop:               //        do {
    add r0, r0, r2   // 1880   sum += prod;
    mov r2, r1       // 1c0a   prod = x;
    mul r2, r1       // 434a   prod *= x;
    sub r1, #1       // 3901   x -= 1;
    mul r2, r1       // 434a   prod *= x;
    bne .loop        // d1f9   } while (prod != 0);
    bx lr            // 4770   return sum;
</code></pre>
<p>A function that takes the input integer via <code>r1</code> and returns the value via <code>r0</code>.</p>
<p>Uses the iterative formula <span class="math-container">\$\sum_{i=1}^{x}{i^2(i-1)}\$</span>, iterating backwards. The loop terminates when the value of <span class="math-container">\$i^2(i-1)\$</span> is 0.</p>
<p>For the purposes of writing test cases, the C-side function signature takes a dummy value for <code>r0</code>.</p>
<p>Test cases written in C:</p>
<pre class="lang-c prettyprint-override"><code>#include &lt;stdio.h&gt;

int func(int dummy, int x);
int main(void) {
    int dummy;
    int inputs[] = {1, 2, 3, 10, 24, 100};
    for(int i = 0; i &lt; 6; ++i) {
        int v = inputs[i];
        printf(&quot;input = %d -&gt; ans = %d\n&quot;, v, func(dummy, v));
    }
    return 0;
}
</code></pre>
<p>Build, run, objdump commands:</p>
<pre class="lang-bash prettyprint-override"><code>arm-linux-gnueabihf-gcc main_c.c func_c.S -static -o main_c
qemu-arm -L /usr/arm-linux-gnueabihf ./main_c
arm-linux-gnueabihf-objdump -d main_c | awk -v RS= '/^[[:xdigit:]]+ &lt;(func|.loop)&gt;/'
</code></pre>
<p>(The objdump + awk trick comes from <a href="https://stackoverflow.com/a/31138400">this SO answer</a>.)</p>


blocks|key|819329|text|维沙尔，8字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|819330|ɾ∑²⁰ɾ²∑-|code-block|syntax|javascript|819331|在网上试试！|unstyled|819332|字面解，平方和，并减去平方和。|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Vyxal/Vyxal|1|https://vyxal.pythonanywhere.com/#WyJQIiwiIiwiyb7iiJHCsuKBsMm%2BwrLiiJEtIiwiIiwiMyJd^0|0|3|0|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|L|3|M|5|K|7|15|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

<h1><a href="https://github.com/Vyxal/Vyxal" rel="nofollow noreferrer">Vyxal</a>, 8 bytes</h1>
<pre><code>ɾ∑²⁰ɾ²∑-
</code></pre>
<p><a href="https://vyxal.pythonanywhere.com/#WyJQIiwiIiwiyb7iiJHCsuKBsMm+wrLiiJEtIiwiIiwiMyJd" rel="nofollow noreferrer">Try it Online!</a></p>
<p>literal solution, squares the sum and subtracts the sum of the squares.</p>


blocks|key|819366|text|因子+%2B+math.unicode，32字节|type|header-two|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|819367|[+[1,b]+3+dupn+Σ+sq+-rot+v.+-+]|code-block|syntax|javascript|819368|在网上试试！|unstyled|819369|word+++%7C+data+stack+++++++++++++++++++++++++++++++++++++%7C+comment
-------%2B------------------------------------------------%2B--------------
+++++++%7C+10+++++++++++++++++++++++++++++++++++++++++++++%7C+example+input
[1,b]++%7C+{+1+2+...+10+}+++++++++++++++++++++++++++++++++%7C
3++++++%7C+{+1+2+...+10+}+3+++++++++++++++++++++++++++++++%7C
dupn+++%7C+{+1+2+...+10+}+{+1+2+...+10+}+{+1+2+...+10+}+++%7C
Σ++++++%7C+{+1+2+...+10+}+{+1+2+...+10+}+55+++++++++++++++%7C+sum
sq+++++%7C+{+1+2+...+10+}+{+1+2+...+10+}+3025+++++++++++++%7C+square
-rot+++%7C+3025+{+1+2+...+10+}+{+1+2+...+10+}+++++++++++++%7C
v.+++++%7C+3025+385+++++++++++++++++++++++++++++++++++++++%7C+dot+product
-++++++%7C+2640+++++++++++++++++++++++++++++++++++++++++++%7C+output|entityMap|0|LINK|mutability|MUTABLE|url|https://factorcode.org/|1|https://tio.run/##JcwxDoJAFIThnlPMAYQAWukBjI2NsSIU6/LEjfB22X2QqPE03scrrRiaKb5J/qvSYn08nw7H/Ra9klvmFbcUcCfP1C00stG2IbTE5FVnnkqM5bCcE/0TAYGGkVjTwnCeRB7OGxbskuSFAiXWKHKUm3lzvJNYoSpWl3rmZnSM7wdhQOqtYMqQoo69csjiDw^0|5|C|0|2|0|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|M|7|15|8|@]|D|@$9|16|A|17|1|18]]|E|$]]|$1|N|3|O|5|H|7|19|8|@]|D|@]|E|$I|J]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

<h1><a href="https://factorcode.org/" rel="nofollow noreferrer">Factor</a> + <code>math.unicode</code>, 32 bytes</h1>
<pre class="lang-factor prettyprint-override"><code>[ [1,b] 3 dupn Σ sq -rot v. - ]
</code></pre>
<p><a href="https://tio.run/##JcwxDoJAFIThnlPMAYQAWukBjI2NsSIU6/LEjfB22X2QqPE03scrrRiaKb5J/qvSYn08nw7H/Ra9klvmFbcUcCfP1C00stG2IbTE5FVnnkqM5bCcE/0TAYGGkVjTwnCeRB7OGxbskuSFAiXWKHKUm3lzvJNYoSpWl3rmZnSM7wdhQOqtYMqQoo69csjiDw" rel="nofollow noreferrer" title="Factor – Try It Online">Try it online!</a></p>
<pre><code>word   | data stack                                     | comment
-------+------------------------------------------------+--------------
       | 10                                             | example input
[1,b]  | { 1 2 ... 10 }                                 |
3      | { 1 2 ... 10 } 3                               |
dupn   | { 1 2 ... 10 } { 1 2 ... 10 } { 1 2 ... 10 }   |
Σ      | { 1 2 ... 10 } { 1 2 ... 10 } 55               | sum
sq     | { 1 2 ... 10 } { 1 2 ... 10 } 3025             | square
-rot   | 3025 { 1 2 ... 10 } { 1 2 ... 10 }             |
v.     | 3025 385                                       | dot product
-      | 2640                                           | output
</code></pre>


blocks|key|823490|text|图2，9\log_{256}(96)\approx+7.408字节|type|header-two|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|823491|a
R@-mQJS|code-block|syntax|javascript|823492|在网上试试！|unstyled|823493|05AB1E港文字解决方案要长2个字符。|823494|a
R@-mQJS
-------
a+++++++#+Range+[1,+n]
-------+#+Use+%5E+as+the+input+to+the+next+line:
+++++JS+#+Prepend+the+sum+to+the+list
+++mQ+++#+Square+each
R@++++++#+Reduce+by
++-+++++#+Subtraction|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Seggan/Fig|1|INLINETEX|IMMUTABLE|teX|9\log_{256}(96)\approx|2|https://fig.fly.dev/#WyJhXG5SQC1tUUpTIiwiMTAiXQ==^0|3|M|0|2|0|3|M|1|0|0|0|6|2|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@$9|18|A|19|1|1A]|$9|1B|A|1C|1|1D]]|E|$]]|$1|F|3|G|5|H|7|1E|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|M|7|1F|8|@]|D|@$9|1G|A|1H|1|1I]]|E|$]]|$1|N|3|O|5|M|7|1J|8|@]|D|@]|E|$]]|$1|P|3|Q|5|H|7|1K|8|@]|D|@]|E|$I|J]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|Z|U|10|E|$11|12]]|13|$5|T|U|V|E|$W|14]]]]

<h1><a href="https://github.com/Seggan/Fig" rel="nofollow noreferrer">Fig</a>, <span class="math-container">\$9\log_{256}(96)\approx\$</span> 7.408 bytes</h1>
<pre><code>a
R@-mQJS
</code></pre>
<p><a href="https://fig.fly.dev/#WyJhXG5SQC1tUUpTIiwiMTAiXQ==" rel="nofollow noreferrer">Try it online!</a></p>
<p>Port of 05AB1E. Literal solution is 2 chars longer.</p>
<pre><code>a
R@-mQJS
-------
a       # Range [1, n]
------- # Use ^ as the input to the next line:
     JS # Prepend the sum to the list
   mQ   # Square each
R@      # Reduce by
  -     # Subtraction
</code></pre>


blocks|key|819589|text|朱莉娅1.0，20字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|819590|使用同大一弦公式：|unstyled|819591|~n=(n%5E3-n)*(3n%2B2)/12|code-block|syntax|javascript|819592|在网上试试！|819593|朴素方法(34个字节)：|819594|~n=(a=[1:n...];sum(a)%5E2-sum(a.%5E2))|819595|entityMap|0|LINK|mutability|MUTABLE|url|http://julialang.org/|1|https://codegolf.stackexchange.com/a/210250/114865|2|https://tio.run/##yyrNyUw0rPj/vy7PViMvzlg3T1NLwzhP20hT39Dof2lxZl66QkBRZl5JGle0QwGYoaGkalxspaCaGZOnpKNQoaNQV6GpkJZfpFChkJmnEG2oY6RjrGNooGNkAiQNYmP/AwA|3|https://tio.run/##yyrNyUw0rPj/vy7PViPRNtrQKk9PTy/Wurg0VyNRM85IF8zQizPS1PxfWpyZl64QUJSZV5LGFe1QAGZoKKkaF1spqGbG5CnpKFToKNRVaCqk5RcpVChk5ilEG@oY6RjrGBroGJkASYPY2P8A^0|0|6|0|0|2|6|1|0|0|0|6|2|0|0|0|0|6|3^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|D|3|E|5|F|7|19|8|@]|9|@$A|1A|B|1B|1|1C]]|C|$]]|$1|G|3|H|5|I|7|1D|8|@]|9|@]|C|$J|K]]|$1|L|3|M|5|F|7|1E|8|@]|9|@$A|1F|B|1G|1|1H]]|C|$]]|$1|N|3|O|5|F|7|1I|8|@]|9|@]|C|$]]|$1|P|3|Q|5|I|7|1J|8|@]|9|@]|C|$J|K]]|$1|R|3|M|5|F|7|1K|8|@]|9|@$A|1L|B|1M|1|1N]]|C|$]]]|S|$T|$5|U|V|W|C|$X|Y]]|Z|$5|U|V|W|C|$X|10]]|11|$5|U|V|W|C|$X|12]]|13|$5|U|V|W|C|$X|14]]]]

<h1><a href="http://julialang.org/" rel="nofollow noreferrer">Julia 1.0</a>, 20 bytes</h1>
<p>Using the <a href="https://codegolf.stackexchange.com/a/210250/114865">same formula as bigyihsuan</a>:</p>

<pre class="lang-julia prettyprint-override"><code>~n=(n^3-n)*(3n+2)/12
</code></pre>
<p><a href="https://tio.run/##yyrNyUw0rPj/vy7PViMvzlg3T1NLwzhP20hT39Dof2lxZl66QkBRZl5JGle0QwGYoaGkalxspaCaGZOnpKNQoaNQV6GpkJZfpFChkJmnEG2oY6RjrGNooGNkAiQNYmP/AwA" rel="nofollow noreferrer" title="Julia 1.0 – Try It Online">Try it online!</a></p>
<hr>
Naive approach (34 bytes):

<pre class="lang-julia prettyprint-override"><code>~n=(a=[1:n...];sum(a)^2-sum(a.^2))
</code></pre>
<p><a href="https://tio.run/##yyrNyUw0rPj/vy7PViPRNtrQKk9PTy/Wurg0VyNRM85IF8zQizPS1PxfWpyZl64QUJSZV5LGFe1QAGZoKKkaF1spqGbG5CnpKFToKNRVaCqk5RcpVChk5ilEG@oY6RjrGBroGJkASYPY2P8A" rel="nofollow noreferrer" title="Julia 1.0 – Try It Online">Try it online!</a></p>


blocks|key|823652|text|Excel，21字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|823653|=(A1%5E3-A1)*(A1/4%2B1/6)|code-block|syntax|javascript|823654|这是别人计算出来的公式的无情的Excel版本。还有一个33字节的解决方案，感觉更像Excel：|unstyled|823655|=LET(n,SEQUENCE(A1),SUM(n%5E3-n%5E2))|823656|..。这个38字节的解决方案看起来最像问题中的原始公式：|823657|=LET(n,SEQUENCE(A1),SUM(n)%5E2-SUMSQ(n))|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|S|8|@]|9|@]|A|$]]|$1|J|3|K|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|L|3|M|5|I|7|U|8|@]|9|@]|A|$]]|$1|N|3|O|5|D|7|V|8|@]|9|@]|A|$E|F]]]|P|$]]

<h1>Excel, 21 bytes</h1>
<pre><code>=(A1^3-A1)*(A1/4+1/6)
</code></pre>
<p>This is the heartless Excel version of the formula figured out by others. There is also the 33 byte solution that feels a little more Excel-y:</p>
<pre><code>=LET(n,SEQUENCE(A1),SUM(n^3-n^2))
</code></pre>
<p>... and the 38 byte solution that looks the most like the original formulas in the question:</p>
<pre><code>=LET(n,SEQUENCE(A1),SUM(n)^2-SUMSQ(n))
</code></pre>


blocks|key|817482|text|APL，13个字符，26个字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|817483|{%2B/⍵×⍵×⍵-1}∘⍳|code-block|syntax|javascript|817484|使用和‘w=1.n’(ww(w-1))的公式(ww(w-1))，可能我写的是其他一些写%2B或-作为“1⊥⍳×⍳×⍳-1”的公式；测试：|unstyled|817485|++g←{%2B/⍵×⍵×⍵-1}∘⍳
++g+0
0
++g+1
0
++g+2
4
++g+3
22
++g+10
2640|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|O|8|@]|9|@]|A|$]]|$1|J|3|K|5|D|7|P|8|@]|9|@]|A|$E|F]]]|L|$]]

<h1>APL(NARS), 13 chars, 26 bytes</h1>
<pre><code>{+/⍵×⍵×⍵-1}∘⍳
</code></pre>
<p>use the formula Sum'w=1..n'(w<em>w</em>(w-1)) possible i wrote the same some other wrote + or - as &quot;1⊥⍳×⍳×⍳-1&quot;; test:</p>
<pre><code>  g←{+/⍵×⍵×⍵-1}∘⍳
  g 0
0
  g 1
0
  g 2
4
  g 3
22
  g 10
2640
</code></pre>


blocks|key|821751|text|斯塔克斯，4+字节数|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|821752|╡⌠(♠|code-block|syntax|javascript|821753|运行并调试它|unstyled|821754|对于输入之前的所有正数k整数，添加k%5E2+*+(k-1)。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/tomtheisen/stax|1|https://github.com/tomtheisen/stax/blob/master/docs/packed.md#packed-stax|2|https://staxlang.xyz/#p=b5f42806&i=1%250A2%250A3%250A10%250A24%250A100&a=1&m=2^0|0|4|0|7|3|1|0|0|0|6|2|0|B|1|H|B^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]|$A|14|B|15|1|16]]|C|$]]|$1|D|3|E|5|F|7|17|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|18|8|@]|9|@$A|19|B|1A|1|1B]]|C|$]]|$1|L|3|M|5|K|7|1C|8|@$A|1D|B|1E|N|O]|$A|1F|B|1G|N|O]]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]|Y|$5|R|S|T|C|$U|Z]]]]

<h1><a href="https://github.com/tomtheisen/stax" rel="nofollow noreferrer">Stax</a>, 4 <a href="https://github.com/tomtheisen/stax/blob/master/docs/packed.md#packed-stax" rel="nofollow noreferrer">bytes</a></h1>

<pre><code>╡⌠(♠
</code></pre>

<p><a href="https://staxlang.xyz/#p=b5f42806&amp;i=1%0A2%0A3%0A10%0A24%0A100&amp;a=1&amp;m=2" rel="nofollow noreferrer">Run and debug it</a></p>

<p>For all positive <code>k</code> integers up to the input, add <code>k^2 * (k-1)</code>.</p>


blocks|key|818396|text|狂暴+-nii，32字节|type|header-two|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|818397|{}{{}[<><<>()>]<([])[][]>}[<>()]|code-block|syntax|javascript|818398|在网上试试！|unstyled|818399|使用堆栈高度遍历k=0+..+n-1，每次迭代都添加k(k%2B1)%5E2。|818400|是如何工作的|818401|main+=+n+add-next-term+0

//+Initial+height+is+0
//+Increase+the+height+at+(push+height),+so+subsequent+`height`+calls+give+k%2B1
add-next-term+=+\x.+x+succ+<(push+height)+height+height>
=+\x.+x+%2B+k+*+(k%2B1)+*+(k%2B1)|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Reconcyl/flurry|1|https://bubbler-4.github.io/Flurry/#!ewB9AHsAewB9AFsAPAA%2BADwAPAA%2BACgAKQA%2BAF0APAAoAFsAXQApAFsAXQBbAF0APgB9AFsAPAA%2BACgAKQBdAA==#MQAwAA==##bgBpAGkA#MQAwADAAMAAwAA==^0|3|4|0|2|0|0|0|0|6|1|0|8|A|Q|8|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|M|7|19|8|@]|D|@$9|1A|A|1B|1|1C]]|E|$]]|$1|N|3|O|5|M|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1I|8|@]|D|@]|E|$]]|$1|R|3|S|5|H|7|1J|8|@]|D|@]|E|$I|J]]]|T|$U|$5|V|W|X|E|$Y|Z]]|10|$5|V|W|X|E|$Y|11]]]]

<h1><a href="https://github.com/Reconcyl/flurry" rel="nofollow noreferrer">Flurry</a> <code>-nii</code>, 32 bytes</h1>
<pre><code>{}{{}[&lt;&gt;&lt;&lt;&gt;()&gt;]&lt;([])[][]&gt;}[&lt;&gt;()]
</code></pre>
<p><a href="https://bubbler-4.github.io/Flurry/#!ewB9AHsAewB9AFsAPAA+ADwAPAA+ACgAKQA+AF0APAAoAFsAXQApAFsAXQBbAF0APgB9AFsAPAA+ACgAKQBdAA==#MQAwAA==##bgBpAGkA#MQAwADAAMAAwAA==" rel="nofollow noreferrer">Try it online!</a></p>
<p>Iterates through <code>k=0 .. n-1</code> using the stack height, adding <code>k(k+1)^2</code> each iteration.</p>
<h3>How it works</h3>
<pre class="lang-haskell prettyprint-override"><code>main = n add-next-term 0

// Initial height is 0
// Increase the height at (push height), so subsequent `height` calls give k+1
add-next-term = \x. x succ &lt;(push height) height height&gt;
= \x. x + k * (k+1) * (k+1)
</code></pre>


blocks|key|818809|text|Arn，9个个字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|818810|└V0¯„○aÌ$|code-block|syntax|javascript|818811|试试看！|unstyled|818812|解释了|818813|解压：%2B{%5E3-%5E2}\~.|style|CODE|818814|执行\large\sum_{k=1}%5E{n}{(k%5E3-k%5E2)}，其中n是输入|818815|++++++++++\+++++++Fold+with
%2B+++++++++++++++++Addition
++{+++++++++++++++After+mapping+with+this+block
++++++++_+++++++++Implied+variable+(implicit+index+of+block).
++++++%5E+++++++++++To+the+power+of
++++++++3+++++++++Literal+three
++++-+++++++++++++Minus
++++++++_
++++++%5E
++++++++2+++++++++Literal+two
++}+++++++++++++++End+block
++++++++++++~+++++1-range+to
++++++++++++++_+++Initialized+to+STDIN;+implied|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/ZippyMagician/Arn|1|https://github.com/ZippyMagician/Arn/wiki/Carn|2|https://zippymagician.github.io/Arn?code=K3teMy1eMn1cfg==&input=MQoyCjMKMTAKMjQ=|3|INLINETEX|IMMUTABLE|teX|\large\sum_{k=1}%5E{n}{(k%5E3-k%5E2)}^0|0|3|0|4|5|1|0|0|0|4|2|0|0|3|A|0|2|V|2|V|3|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@$A|1C|B|1D|1|1E]|$A|1F|B|1G|1|1H]]|C|$]]|$1|D|3|E|5|F|7|1I|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1J|8|@]|9|@$A|1K|B|1L|1|1M]]|C|$]]|$1|L|3|M|5|6|7|1N|8|@]|9|@]|C|$]]|$1|N|3|O|5|K|7|1O|8|@$A|1P|B|1Q|P|Q]]|9|@]|C|$]]|$1|R|3|S|5|K|7|1R|8|@$A|1S|B|1T|P|Q]]|9|@$A|1U|B|1V|1|1W]]|C|$]]|$1|T|3|U|5|F|7|1X|8|@]|9|@]|C|$G|H]]]|V|$W|$5|X|Y|Z|C|$10|11]]|12|$5|X|Y|Z|C|$10|13]]|14|$5|X|Y|Z|C|$10|15]]|16|$5|17|Y|18|C|$19|1A]]]]

<h1><a href="https://github.com/ZippyMagician/Arn" rel="nofollow noreferrer">Arn</a>, <a href="https://github.com/ZippyMagician/Arn/wiki/Carn" rel="nofollow noreferrer">9 bytes</a></h1>
<pre><code>└V0¯„○aÌ$
</code></pre>
<p><a href="https://zippymagician.github.io/Arn?code=K3teMy1eMn1cfg==&amp;input=MQoyCjMKMTAKMjQ=" rel="nofollow noreferrer">Try it!</a></p>
<h1>Explained</h1>
<p>Unpacked: <code>+{^3-^2}\~</code>.</p>
<p>Performs <span class="math-container">\$\large\sum_{k=1}^{n}{(k^3-k^2)}\$</span> where n is the input</p>
<pre><code>          \       Fold with
+                 Addition
  {               After mapping with this block
        _         Implied variable (implicit index of block).
      ^           To the power of
        3         Literal three
    -             Minus
        _
      ^
        2         Literal two
  }               End block
            ~     1-range to
              _   Initialized to STDIN; implied
</code></pre>


blocks|key|822756|text|Haskell，22字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|822757|f+n=(3*n%2B2)*(n%5E3-n)/12|code-block|syntax|javascript|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|I|8|@]|9|@]|A|$E|F]]]|G|$]]

<h1>Haskell, 22 bytes</h1>
<pre><code>f n=(3*n+2)*(n^3-n)/12
</code></pre>


blocks|key|823226|text|C+(gcc)，29字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|823227|f(n){n=(3*n%2B2)*(n*n*n-n)/12;}|code-block|syntax|javascript|823228|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://gcc.gnu.org/|1|https://tio.run/##S9ZNT07@/z9NI0@zOs9Ww1grT9tIU0sjTwsIdfM09Q2NrGv/5yZm5mloVnMVFGXmlaRpKKmmxOQp6aRpGGpqWmMIGmETNMYmaGiAVb8JdrVgxbX/AQ^0|0|7|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<h1><a href="https://gcc.gnu.org/" rel="nofollow noreferrer">C (gcc)</a>, 29 bytes</h1>

<pre class="lang-c prettyprint-override"><code>f(n){n=(3*n+2)*(n*n*n-n)/12;}
</code></pre>
<p><a href="https://tio.run/##S9ZNT07@/z9NI0@zOs9Ww1grT9tIU0sjTwsIdfM09Q2NrGv/5yZm5mloVnMVFGXmlaRpKKmmxOQp6aRpGGpqWmMIGmETNMYmaGiAVb8JdrVgxbX/AQ" rel="nofollow noreferrer" title="C (gcc) – Try It Online">Try it online!</a></p>


blocks|key|819487|text|J，15字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|819488|(%2B/@:**:@>:)@i.|code-block|syntax|javascript|819489|使用表单\sum_{i=0}%5E{n-1}((i%2B1)%5E2\cdot+i)|unstyled|style|CODE|819490|在网上试试！|819491|(%2B/@:**:@>:)@i.
+++++++++++++i.+NB.+range+0..n-1
++++++++++++@+++NB.+atop+f+(g+x)
(++++++++++)++++NB.+monadic+hook,+y+f+(g+y)
++++++*:@>:+++++NB.+increment+then+square
+%2B/@:*++++++++++NB.+multiply+square+with+each+i+then+sum+the+result
++++++++++++++++NB.+u@:v+where+v+is+dyadic+and+u+is+always+monadic,
++++++++++++++++NB.+like+atop+except+u+will+execute+once+on+the+entire+result|entityMap|0|LINK|mutability|MUTABLE|url|https://www.jsoftware.com/|1|INLINETEX|IMMUTABLE|teX|\sum_{i=0}%5E{n-1}((i%2B1)%5E2\cdot+i)|2|https://ato.pxeger.com/run?1=m70wa8FOJT31NAVbKwV1BR0FAwUrINbVU3AO8nFbWlqSpmuxXkNb38FKS8vKwc5K0yFTDyK6S5MrNTkjXyFNzU7BUMFIwVjB0EDByARIGkAULFgAoQE^0|0|1|0|0|0|4|W|4|W|1|0|0|6|2|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|D|3|E|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1A|8|@$A|1B|B|1C|L|M]]|9|@$A|1D|B|1E|1|1F]]|C|$]]|$1|N|3|O|5|K|7|1G|8|@]|9|@$A|1H|B|1I|1|1J]]|C|$]]|$1|P|3|Q|5|F|7|1K|8|@]|9|@]|C|$G|H]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|Z|U|10|C|$11|12]]|13|$5|T|U|V|C|$W|14]]]]

<h1><a href="https://www.jsoftware.com" rel="nofollow noreferrer">J</a>, 15 bytes</h1>
<pre><code>(+/@:**:@&gt;:)@i.
</code></pre>
<p>Uses the form <span class="math-container">\$\sum_{i=0}^{n-1}((i+1)^2\cdot i)\$</span></p>
<p><a href="https://ato.pxeger.com/run?1=m70wa8FOJT31NAVbKwV1BR0FAwUrINbVU3AO8nFbWlqSpmuxXkNb38FKS8vKwc5K0yFTDyK6S5MrNTkjXyFNzU7BUMFIwVjB0EDByARIGkAULFgAoQE" rel="nofollow noreferrer">Attempt This Online!</a></p>
<pre><code>(+/@:**:@&gt;:)@i.
             i. NB. range 0..n-1
            @   NB. atop f (g x)
(          )    NB. monadic hook, y f (g y)
      *:@&gt;:     NB. increment then square
 +/@:*          NB. multiply square with each i then sum the result
                NB. u@:v where v is dyadic and u is always monadic,
                NB. like atop except u will execute once on the entire result
</code></pre>


blocks|key|819681|text|维沙尔，7字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|819682|ɾ:∑p²ƒ-|code-block|syntax|javascript|819683|在网上试试！|unstyled|819684|图中的一个港口，是05ab1e号的港口|819685|解释了|819686|ɾ:∑p²ƒ-
ɾ:∑p+++++#+prepend+the+sum+of+the+range+[1,+n]+to+the+range+[1,+n]
+++²ƒ-++#+square+everything+and+reduce+by+subtraction|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Vyxal/Vyxal|1|https://vyxal.pythonanywhere.com/#WyJBIiwiIiwiyb464oiRcMKyxpItIiwiIiwiMVxuMlxuM1xuMTBcbjI0XG4xMDAiXQ==^0|0|3|0|0|0|0|6|1|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|L|3|M|5|K|7|19|8|@]|9|@]|C|$]]|$1|N|3|O|5|6|7|1A|8|@]|9|@]|C|$]]|$1|P|3|Q|5|F|7|1B|8|@]|9|@]|C|$G|H]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|T|U|V|C|$W|Z]]]]

<h1><a href="https://github.com/Vyxal/Vyxal" rel="nofollow noreferrer">Vyxal</a>, 7 bytes</h1>
<pre><code>ɾ:∑p²ƒ-
</code></pre>
<p><a href="https://vyxal.pythonanywhere.com/#WyJBIiwiIiwiyb464oiRcMKyxpItIiwiIiwiMVxuMlxuM1xuMTBcbjI0XG4xMDAiXQ==" rel="nofollow noreferrer">Try it Online!</a></p>
<p>A port of Fig which is a port of 05ab1e</p>
<h2>Explained</h2>
<pre><code>ɾ:∑p²ƒ-
ɾ:∑p     # prepend the sum of the range [1, n] to the range [1, n]
   ²ƒ-  # square everything and reduce by subtraction
</code></pre>


blocks|key|821615|text|公理，39字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|821616|f(n)==reduce(%2B,[x%5E3-x%5E2+for+x+in+1..n])|code-block|syntax|javascript|821617|测试：|unstyled|821618|->+[[x,f+x]for+x+in+[1,2,3,10]]
+++++[[1,0],[2,4],[3,22],[10,2640]]|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|O|8|@]|9|@]|A|$]]|$1|J|3|K|5|D|7|P|8|@]|9|@]|A|$E|F]]]|L|$]]

<h1>Axiom, 39 bytes</h1>
<pre><code>f(n)==reduce(+,[x^3-x^2 for x in 1..n])
</code></pre>
<p>test:</p>
<pre><code>-&gt; [[x,f x]for x in [1,2,3,10]]
     [[1,0],[2,4],[3,22],[10,2640]]
</code></pre>


blocks|key|821862|text|克洛尔，91字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|821863|(fn+s[n](let[u+#(apply+%2B+%25)q+#(Math/pow+%25+2)m(range+1(inc+n))](-(q(u+m))(u(map+#(q+%25)m)))))|code-block|syntax|javascript|821864|天真的直截了当的方法。见下文：|unstyled|821865|(defn+sum-sq-diff+[n]
++++(let+[;+Shortcut+functions+to+save+bytes
++++++++++sum+#(apply+%2B+%25)
++++++++++square+#(Math/pow+%25+2)
++++++++++nums+(range+1+(inc+n))

++++++++++ss1+(sum+(map+#(square+%25)+nums))
++++++++++ss2+(square+(sum+nums))]

++++++(-+ss2+ss1)))

(mapv+sum-sq-diff+[1+2+3+10+24+100])
=>+[0.0+4.0+22.0+2640.0+85100.0+2.516415E7]|821866|在网上试试！|entityMap|0|LINK|mutability|MUTABLE|url|https://clojure.org/|1|https://tio.run/##DY1BCsIwFAWv8kAK7yOidevKA3iC0kUosa0kv2maIJ4@/tnNYpgpbJ@afWPKq5agYONbcQw6MvgyVJzoUgo/nNHJbvZyZbmm7YsOd4nMTmePnqtOUJGRF@6siCKsjC5ZsVtpbrT@JoIHnnnG4m37Bw^0|0|3|0|0|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|13|8|@]|9|@]|C|$]]|$1|L|3|M|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|N|3|O|5|K|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

<h1><a href="https://clojure.org/" rel="nofollow noreferrer">Clojure</a>, 91 bytes</h1>



<pre class="lang-clj prettyprint-override"><code>(fn s[n](let[u #(apply + %)q #(Math/pow % 2)m(range 1(inc n))](-(q(u m))(u(map #(q %)m)))))
</code></pre>

<p>The naive, literal approach. See below:</p>

<pre class="lang-clj prettyprint-override"><code>(defn sum-sq-diff [n]
    (let [; Shortcut functions to save bytes
          sum #(apply + %)
          square #(Math/pow % 2)
          nums (range 1 (inc n))

          ss1 (sum (map #(square %) nums))
          ss2 (square (sum nums))]

      (- ss2 ss1)))

(mapv sum-sq-diff [1 2 3 10 24 100])
=&gt; [0.0 4.0 22.0 2640.0 85100.0 2.516415E7]
</code></pre>

<p><a href="https://tio.run/##DY1BCsIwFAWv8kAK7yOidevKA3iC0kUosa0kv2maIJ4@/tnNYpgpbJ@afWPKq5agYONbcQw6MvgyVJzoUgo/nNHJbvZyZbmm7YsOd4nMTmePnqtOUJGRF@6siCKsjC5ZsVtpbrT@JoIHnnnG4m37Bw" rel="nofollow noreferrer" title="Clojure – Try It Online">Try it online!</a></p>


blocks|key|822138|text|红宝石，24字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|822139|->n{(n%2Bn%2B3*n*=n)*~-n/12}|code-block|syntax|javascript|822140|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://www.ruby-lang.org/|1|https://tio.run/##KypNqvyfZhvzX9cur1ojTztP21grT8s2T1OrTjdP39Co9n@0oY6RjrGOoYGOkQmQNIjVy00sqK6pqClQiK7QSYuuiI2t/Q8A^0|0|3|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<h1><a href="https://www.ruby-lang.org/" rel="nofollow noreferrer">Ruby</a>, 24 bytes</h1>



<pre class="lang-ruby prettyprint-override"><code>-&gt;n{(n+n+3*n*=n)*~-n/12}
</code></pre>

<p><a href="https://tio.run/##KypNqvyfZhvzX9cur1ojTztP21grT8s2T1OrTjdP39Co9n@0oY6RjrGOoYGOkQmQNIjVy00sqK6pqClQiK7QSYuuiI2t/Q8A" rel="nofollow noreferrer" title="Ruby – Try It Online">Try it online!</a></p>


blocks|key|822212|text|05AB1E，6字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|822213|LDOšnÆ|code-block|syntax|javascript|822214|在网上试试！|unstyled|822215|解释：|822216|+++++++++++#+implicit+input+(example:+3)
L++++++++++#+range+([1,+2,+3])
+DOš+++++++#+prepend+the+sum+([6,+1,+2,+3])
++++n++++++#+square+each+([36,+1,+4,+9])
+++++Æ+++++#+reduce+by+subtraction+(22)
+++++++++++#+implicit+output|822217|Æ不是经常有用的，但现在是它发光的时候了。这比天真的LOnILnO-多了两个字节。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Adriandmen/05AB1E|1|https://tio.run/##yy9OTMpM/f/fx8X/6MK8w23//xsaAAA^0|0|6|0|0|0|0|6|1|0|0|0|0|1|Q|8^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|17|8|@]|9|@$A|18|B|19|1|1A]]|C|$]]|$1|L|3|M|5|K|7|1B|8|@]|9|@]|C|$]]|$1|N|3|O|5|F|7|1C|8|@]|9|@]|C|$G|H]]|$1|P|3|Q|5|K|7|1D|8|@$A|1E|B|1F|R|S]|$A|1G|B|1H|R|S]]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]|10|$5|V|W|X|C|$Y|11]]]]

<h1><a href="https://github.com/Adriandmen/05AB1E" rel="nofollow noreferrer">05AB1E</a>, 6 bytes</h1>

<pre><code>LDOšnÆ
</code></pre>

<p><a href="https://tio.run/##yy9OTMpM/f/fx8X/6MK8w23//xsaAAA" rel="nofollow noreferrer" title="05AB1E – Try It Online">Try it online!</a></p>

<p>Explanation:</p>

<pre><code>           # implicit input (example: 3)
L          # range ([1, 2, 3])
 DOš       # prepend the sum ([6, 1, 2, 3])
    n      # square each ([36, 1, 4, 9])
     Æ     # reduce by subtraction (22)
           # implicit output
</code></pre>

<p><code>Æ</code> isn't useful often, but this is its time to shine. This beats the naïve <code>LOnILnO-</code> by two whole bytes.</p>


blocks|key|822479|text|木桶，15字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|822480|&⑻Ï⑷²⑸⅀⑻:⑨*½²$-|code-block|syntax|javascript|822481|在网上试试！|unstyled|822482|计算平方和，然后计算和的平方。|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/JonoCode9374/Keg|1|https://tio.run/##y05N//9f7dHE3Yf7H03cfmjTo4k7HrU2APlWjyau0Dq099AmFd3//w0NAA^0|0|2|0|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|L|3|M|5|K|7|15|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

<h1><a href="https://github.com/JonoCode9374/Keg" rel="nofollow noreferrer">Keg</a>, 15 bytes</h1>
<pre><code>&amp;⑻Ï⑷²⑸⅀⑻:⑨*½²$-
</code></pre>
<p><a href="https://tio.run/##y05N//9f7dHE3Yf7H03cfmjTo4k7HrU2APlWjyau0Dq099AmFd3//w0NAA" rel="nofollow noreferrer" title="Keg – Try It Online">Try it online!</a></p>
<p>Calculates the sum of squares then the square of the sum.</p>


blocks|key|818629|text|耳语v2，70字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|818630|>+2
>+Input
>>+L³
>>+L²
>>+∑+1+2+3
>>+∑+1+2+4
>>+5-6
>>+Output+7|code-block|syntax|javascript|818631|在网上试试！|unstyled|818632|是如何工作的|818633|实现公式|818634|818635|\sum%5En_{i=2}i%5E3+-+\sum%5En_{i=2}i%5E2|atomic|style|CODE|mathjax|teX|\sum%5En_{i=2}i%5E3+-+\sum%5En_{i=2}i%5E2|818636|818637|由于∑、Each等语句中的复合语句很难处理，因此计算两个和并取其差值的时间实际上比计算短|818638|818639|\sum%5En_{i=2}(i%5E3-i%5E2)|\sum%5En_{i=2}(i%5E3-i%5E2)|818640|818641|或者其他任何一笔钱。|818642|这必须是一个最容易阅读的低语程序，除了非常琐碎的程序。我们使用∑语句计算这两个和，其工作方式是将起始界作为第一个值，将结束界作为第二个值，最后将函数取为最后一个和。这两个∑语句都有相同的界限--+2和n+--所以唯一不同的是函数。对于第一个∑语句，我们使用语句>>+L³+(即多维数据集)和>>+L²+(square)作为第二个语句。这产生了\sum%5En_{i=2}i%5E3和\sum%5En_{i=2}i%5E2之和。最后，我们利用他们的差异并输出它。|818643|耳语v3，48字节|818644|>+2
>+Input
>>>+L³-L²
>>+∑+1+2+3
>>+Output+4|818645|在第3版中引入模式线(从>>>开始)后，Whispers中的复合语句更容易处理。不幸的是，Whispers+v3目前还不能在TIO上使用，所以您必须使用Github中的解释器。|818646|此版本利用模式行缩短代码，只在和中使用一条语句，而不是使用两个单独的和。除此之外，考虑到v2程序，它的工作原理相当清楚。|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/cairdcoinheringaahing/Whispers|1|https://tio.run/##K8/ILC5ILSo2@v/fTsGIy07BM6@gtITLzk7B59BmCLUJRD3qmKhgqGCkYIzMMQFxTHXNQJR/aQlQo4L5//@GBgYA|2|INLINETEX|IMMUTABLE|3|n|4|\sum%5En_{i=2}i%5E3|5|\sum%5En_{i=2}i%5E2|6^0|0|4|0|0|0|0|6|1|0|0|0|0|0|X|0|0|2|1|4|4|0|0|0|L|0|0|0|V|1|2D|1|2Q|1|2S|1|3C|1|3M|5|41|5|4R|F|57|F|2Q|1|2|2S|1|3|4R|F|4|57|F|5|0|0|4|6|0|0|C|3|0^^$0|@$1|2|3|4|5|6|7|21|8|@]|9|@$A|22|B|23|1|24]]|C|$]]|$1|D|3|E|5|F|7|25|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|26|8|@]|9|@$A|27|B|28|1|29]]|C|$]]|$1|L|3|M|5|6|7|2A|8|@]|9|@]|C|$]]|$1|N|3|O|5|K|7|2B|8|@]|9|@]|C|$]]|$1|P|3|-4|5|K|7|2C|8|@]|9|@]|C|$]]|$1|Q|3|R|5|S|7|2D|8|@$A|2E|B|2F|T|U]]|9|@]|C|$V|-1|W|X]]|$1|Y|3|-4|5|K|7|2G|8|@]|9|@]|C|$]]|$1|Z|3|10|5|K|7|2H|8|@$A|2I|B|2J|T|U]|$A|2K|B|2L|T|U]]|9|@]|C|$]]|$1|11|3|-4|5|K|7|2M|8|@]|9|@]|C|$]]|$1|12|3|13|5|S|7|2N|8|@$A|2O|B|2P|T|U]]|9|@]|C|$V|-1|W|14]]|$1|15|3|-4|5|K|7|2Q|8|@]|9|@]|C|$]]|$1|16|3|17|5|K|7|2R|8|@]|9|@]|C|$]]|$1|18|3|19|5|K|7|2S|8|@$A|2T|B|2U|T|U]|$A|2V|B|2W|T|U]|$A|2X|B|2Y|T|U]|$A|2Z|B|30|T|U]|$A|31|B|32|T|U]|$A|33|B|34|T|U]|$A|35|B|36|T|U]|$A|37|B|38|T|U]|$A|39|B|3A|T|U]]|9|@$A|3B|B|3C|1|3D]|$A|3E|B|3F|1|3G]|$A|3H|B|3I|1|3J]|$A|3K|B|3L|1|3M]]|C|$]]|$1|1A|3|1B|5|6|7|3N|8|@]|9|@$A|3O|B|3P|1|3Q]]|C|$]]|$1|1C|3|1D|5|F|7|3R|8|@]|9|@]|C|$G|H]]|$1|1E|3|1F|5|K|7|3S|8|@$A|3T|B|3U|T|U]]|9|@]|C|$]]|$1|1G|3|1H|5|K|7|3V|8|@]|9|@]|C|$]]]|1I|$1J|$5|1K|1L|1M|C|$1N|1O]]|1P|$5|1K|1L|1M|C|$1N|1Q]]|1R|$5|1S|1L|1T|C|$W|1R]]|1U|$5|1S|1L|1T|C|$W|1V]]|1W|$5|1S|1L|1T|C|$W|1X]]|1Y|$5|1S|1L|1T|C|$W|1Z]]|20|$5|1K|1L|1M|C|$1N|1O]]]]

<h1><a href="https://github.com/cairdcoinheringaahing/Whispers" rel="nofollow noreferrer">Whispers v2</a>, 70 bytes</h1>
<pre><code>&gt; 2
&gt; Input
&gt;&gt; L³
&gt;&gt; L²
&gt;&gt; ∑ 1 2 3
&gt;&gt; ∑ 1 2 4
&gt;&gt; 5-6
&gt;&gt; Output 7
</code></pre>
<p><a href="https://tio.run/##K8/ILC5ILSo2@v/fTsGIy07BM6@gtITLzk7B59BmCLUJRD3qmKhgqGCkYIzMMQFxTHXNQJR/aQlQo4L5//@GBgYA" rel="nofollow noreferrer" title="Whispers v2 – Try It Online">Try it online!</a></p>
<h2>How it works</h2>
<p>Implements the formula</p>
<p><span class="math-container">$$\sum^n_{i=2}i^3 - \sum^n_{i=2}i^2$$</span></p>
<p>As Whispers is difficult to deal with when it comes to compound statements inside of <code>∑</code>, <code>Each</code> etc. statements, it's actually shorter to calculate two sums and take their difference than it is to calculate</p>
<p><span class="math-container">$$\sum^n_{i=2}(i^3-i^2)$$</span></p>
<p>or any other single sum.</p>
<p>This has to be one of the easiest-to-read Whispers programs, aside from very trivial ones. We use the <code>∑</code> statement to calculate the two sums, which works by taking the starting bound as the first value, the end bound second and finally the function to sum through last. Both of the <code>∑</code> statements have the same bounds - <span class="math-container">\$2\$</span> and <span class="math-container">\$n\$</span> - so the only thing that differs is the function. For the first <code>∑</code> statement, we use the statement <code>&gt;&gt; L³</code> (i.e. cube), and <code>&gt;&gt; L²</code> (square) for the second. This yields the sums <span class="math-container">\$\sum^n_{i=2}i^3\$</span> and <span class="math-container">\$\sum^n_{i=2}i^2\$</span>. Finally, we take their difference and output it.</p>
<h1><a href="https://github.com/cairdcoinheringaahing/Whispers" rel="nofollow noreferrer">Whispers v3</a>, 48 bytes</h1>
<pre><code>&gt; 2
&gt; Input
&gt;&gt;&gt; L³-L²
&gt;&gt; ∑ 1 2 3
&gt;&gt; Output 4
</code></pre>
<p>With the introduction of pattern lines (starting with <code>&gt;&gt;&gt;</code>) in version 3, compound statements in Whispers are much easier to deal with. Unfortunately, Whispers v3 is currently not available on TIO, so you'll have to use the interpreter in the Github repo.</p>
<p>This version takes advantage of pattern lines to shorten the code by only using one statement in the sum, rather than two separate sums. Other than that, it's fairly clear how it works, given the v2 program.</p>


blocks|key|822971|text|05AB1E，11字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|822972|LDOnU€nOXs-|code-block|syntax|javascript|822973|在网上试试！|unstyled|822974|+++++++++++++#+(implicit)+push+STDIN+to+stack
LD+++++++++++#+push+[1,+2,+3,+...,+top+of+stack]+twice
++OnU++++++++#+store+squared+sum+of+all+values+in+top+of+stack+in+variable+X
+++++€+++++++#+for+each+in+top+of+stack...
++++++n++++++#+push+top+of+stack+squared
+++++++O+++++#+push+sum+of+all+values+in+top+of+stack
++++++++Xs-++#+push+X+-+top+of+stack
+++++++++++++#+(implicit)+output+top+of+stack+to+STDOUT|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Adriandmen/05AB1E|1|https://tio.run/##yy9OTMpM/f/fx8U/L/RR05o8/4hi3f//AQ^0|0|6|0|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|L|3|M|5|F|7|15|8|@]|9|@]|C|$G|H]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

<h1><a href="https://github.com/Adriandmen/05AB1E" rel="nofollow noreferrer">05AB1E</a>, 11 bytes</h1>
<pre><code>LDOnU€nOXs-
</code></pre>
<p><a href="https://tio.run/##yy9OTMpM/f/fx8U/L/RR05o8/4hi3f//AQ" rel="nofollow noreferrer" title="05AB1E – Try It Online">Try it online!</a></p>
<pre><code>             # (implicit) push STDIN to stack
LD           # push [1, 2, 3, ..., top of stack] twice
  OnU        # store squared sum of all values in top of stack in variable X
     €       # for each in top of stack...
      n      # push top of stack squared
       O     # push sum of all values in top of stack
        Xs-  # push X - top of stack
             # (implicit) output top of stack to STDOUT
</code></pre>


<p>The sum of the squares of the first ten natural numbers is,
<span class="math-container">\$1^2 + 2^2 + \dots + 10^2 = 385\$</span></p>

<p>The square of the sum of the first ten natural numbers is,</p>

<p><span class="math-container">\$(1 + 2 + ... + 10)^2 = 55^2 = 3025\$</span></p>

<p>Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is </p>

<p><span class="math-container">\$3025 − 385 = 2640\$</span></p>

<p>For a given input n, find the difference between the sum of the squares of the first n natural numbers and the square of the sum.</p>

<p>Test cases</p>

<pre><code>1       =&gt; 0
2       =&gt; 4
3       =&gt; 22
10      =&gt; 2640
24      =&gt; 85100
100     =&gt; 25164150
</code></pre>

<p>This challenge was first announced at <a href="https://projecteuler.net/problem=6" rel="noreferrer">Project Euler #6</a>.</p>

<h3>Winning Criteria</h3>

<ul>
<li><p>There are no rules about what should be the behavior with negative or zero input.</p></li>
<li><p>The shortest answer wins.</p></li>
</ul>


Sum square difference

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前十个自然数的平方之和是，1^2 + 2^2 + \dots + 10^2 = 385前十个自然数之和的平方是，(1 + 2 + ... + 10)^2 = 55^2 = 3025因此，前十个自然数的平方和与和的平方之间的差额是3025 − 385 = 2640对于给定的输入n，求出第一个n个自然数的平方和与和的平方之间的差。测试用例1       => 02       => 43       

问和平方差
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