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?
Only issue to me is that you should not be saying $\frac{(k+1)^2(k+2)^2}{4} = P(k+1)$. $P$ is a predicate, not a polynomial. In other words, it is a mathematical statement which is either true or false, but it does not possess a numerical value, unlike the arithmetic expression given on the left side. Thus you might instead write $1^3 + 2^3 + \dots + (k+1)^3 = \frac{(k+1)^2(k+2)^2}{4} \implies P(k+1)$.
Having logical connectives (e.g., $\implies$)between each of your equations is probably a good idea as well, because even though your reader likely recognizes that these statements are meant to be taken together, in theory they could just be unrelated statements.