<s>[INST] How to determine equation for $\sum_{k=1}^n k^3$ How do you find an algebraic formula for $\sum_{k=1}^n k^3$? I am able to find one for $\sum_{k=1}^n k^2$, but not $k^3$. Any hints would be appreciated.
 [/INST] One last Proof :
$$\sum_{k=1}^{n}k^{3}=1+8+27+64+...+n^{3}$$
$$\sum_{k=1}^{n}k^{3}=\underbrace{1}_{1^{3}}+\underbrace{3+5}_{2^{3}}+\underbrace{7+9+11}_{3^{3}}+\cdots+\underbrace{\left(n(n-1)+1\right)+\cdots+\left(n(n+1)-1\right)}_{n^{3}}$$
$$\sum_{k=1}^{n}k^{3}=1+3+5+\cdots+\left(n(n+1)-1\right)$$
Since $n(n+1) -1 $ is odd, then $n(n+1) -1 $ is the form $2k -1$
$$n(n+1) -1 = 2k -1 $$
$$n(n+1) = 2k $$
$$\underbrace{\frac{n}{2}(n+1)}_{\text{Arithmetic progression}} = k $$
$$\sum_{k=1}^{n}k^{3}=\underbrace{\underbrace{\underbrace{\underbrace{1}_{1^{2}}+3}_{2^{2}}+5}_{3^{2}}+\cdots+\left(n(n+1)-1\right)}_{ \left(\frac{n^{2}+n}{2}\right)^{2}}$$

$$\sum_{k=1}^{n}k^{3}=\left(\frac{n^{2}+n}{2}\right)^{2}=\frac{1}{4}n^2(n+1)^2$$

Note: 
$\sum_{k=1}^{n}(2k-1)=1+3+5+7+9+...+2n-1=n^{2}$
</s>