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$$\sum_{k=1}^n {(k^3)} = \left(\sum_{k=1}^n {k}\right)^2 $$

How could I approach this problem to make a combinatorial proof?

Brian M. Scott
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Jace
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  • A combinatorial sum related:$$\sum_{k=1}^n\frac{k(k+1)(k+2)\dots(k+p-1)}{p!}=\frac{n(n+1)(n+2)(n+3)\dots(n+p)}{(p+1)!}$$ which is really saying $$\sum_{k=1}^n\binom{k+p-1}p=\binom{k+p}{p+1}$$Exploit the cases $p=3,2,1$ – Simply Beautiful Art Nov 11 '16 at 00:40

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Try induction. Suppose the statement holds for $n-1$. To prove it for $n$, write: $$ \left( \sum_{k=1}^{n} k \right)^2 = \left( n + \sum_{k=1}^{n-1} k \right)^2 = n^2 + 2 n \sum_{k=1}^{n-1} k + \left( \sum_{k=1}^{n-1} k \right)^2 = n^2 + 2 n \sum_{k=1}^{n-1} k + \sum_{k=1}^{n-1} k^3. $$ Now use $$ \sum_{k=1}^{n-1} k = {(n-1) \, n \over 2}. $$

avs
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