<s>[INST] Summation Identity: $\sum_{i=1}^ni^3 = \left( \frac{n(n+1)}{2} \right)^2$ I have to prove:
$$\sum\limits_{i = 1}^n i^3 = \Bigg( \frac{n(n+1)}{2}\Bigg)^2$$
Using the following:
$$n^3 = 6 {n \choose 3} + 6 {n \choose 2} + n \quad \forall n \in \mathbb{N}$$
My work is that first I substitute $n^3$ for $6 {n \choose 3} + 6 {n \choose 2} + n$. Then I go and invoke the sum over that (I am assuming this is how it works). That is,
$$\sum\limits^n_{i=1} \bigg(6 {i \choose 3} + 6 {i \choose 2} + i \bigg)$$
$$6 \sum\limits^n_{i=1} {i \choose 3} + 6 \sum\limits^n_{i=1} {i \choose 2} + \sum\limits^n_{i=1} {i \choose 1}$$
The summation identity is:
$$\sum\limits^n_{i=0} {i \choose k} = {n+1 \choose k+1}$$
invoking it over the sums yields:
$$6{n+1 \choose 4} + 6{n+1 \choose 3} + {n+1 \choose 2}$$
I think these are the right steps, but simplification seems a bit difficult to get the ending result.
 [/INST] It’s just algebra $-$ tedious, but straightforward:
$$\begin{align*}
6\binom{n+1}4&+6\binom{n+1}3+\binom{n+1}2\\
\strut\\
&=\frac{6(n+1)n(n-1)(n-2)}{4!}+\frac{6(n+1)n(n-1)}{3!}+\frac{(n+1)n}2\\
&=n(n+1)\left(\frac{(n-1)(n-2)}{4}+(n-1)+\frac12\right)\\
&=\frac{n(n+1)}4\big((n-1)(n-2)+4(n-1)+2\big)\\
&=\frac{n(n+1)}4\big((n-1)(n+2)+2\big)\\
&=\frac{n(n+1)}4(n^2+n)\\
&=\frac{n^2(n+1)^2}4\\
&=\left(\frac{n(n+1)}2\right)^2
\end{align*}$$
</s>