<s>[INST] Induction proof that $1^3+2^3+...+n^3=\frac{n^2(n+1)^2}4$ Prove that:
$1^3+2^3+...+n^3=\frac{n^2(n+1)^2}{4}$ for $n \in N$
So I am thinking that I need to do a proof by mathematical induction. Here's my attempt:
Let S(n) be the statement $1^3+2^3+...+n^3=\frac{n^2(n+1)^2}{4}$ for $n \in N$. When n=1, 
$$1^3=\frac{1^2(1+1)^2}{4}$$
$$1=\frac{(2)^2}{4}$$
$$1=\frac{4}{4}$$
So P(1) is true.
Suppose that k is any integer with $k \ge 1$ such that  $p(k)1^3+2^3+...+k^3=\frac{k^2(k+1)^2}{4}$ for $k \in N$. Then adding $(k+1)^3$ to both sides,
$$1^3+2^3+...+k^3=\frac{k^2(k+1)^2}{4}$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{k^2(k+1)^2}{4}+(k+1)^3$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{k^2(k+1)^2+4(k+1)^3}{4}$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k^2+4(k+1)}{4}$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k^2+4k+4)}{4}$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4}$$
$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k+1+1)^2}{4}=P(k+1)$$
Thus, since we have proved that P(k+1) is true, we can conclude that P(k) is true for all $k \in N$.
How does it look, any edits I should make?
 [/INST] Ok, It is good. 
I would do it as this. In the first step for $(k+1)$,
Left side:
$$P_{k+1} - P_k = (k+1)^3$$
Right side:
$$P_{k+1}-P_k={{(k+1)^2}{(k+2)^2} \over 4}-{{(k)^2}{(k+1)^2} \over 4}$$
$$P_{k+1}-P_k={1 \over 4}{(k+1)^2}({(k+2)^2-(k)^2})$$
$$P_{k+1}-P_k={1 \over 4}{(k+1)^2}({4k+4})$$
$$P_{k+1}-P_k={(k+1)^3}$$
Therefor:
$L=R$
</s>