Prove by induction $\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$ for $n\ge1$ Prove the following statement $S(n)$ for $n\ge1$:
$$\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$$
To prove the basis, I substitute $1$ for $n$ in $S(n)$:
$$\sum_{i=1}^11^3=1=\frac{1^2(2)^2}{4}$$
Great. For the inductive step, I assume $S(n)$ to be true and prove $S(n+1)$:
$$\sum_{i=1}^{n+1}i^3=\frac{(n+1)^2(n+2)^2}{4}$$
Considering the sum on the left side:
$$\sum_{i=1}^{n+1}i^3=\sum_{i=1}^ni^3+(n+1)^3$$
I make use of $S(n)$ by substituting its right side for $\sum_{i=1}^ni^3$:
$$\sum_{i=1}^{n+1}i^3=\frac{n^2(n+1)^2}{4}+(n+1)^3$$
This is where I get a little lost. I think I expand the equation to be
$$=\frac{(n^4+2n^3+n^2)}{4}+(n+1)^3$$
but I'm not totally confident about that. Can anyone provide some guidance?

Here's yet another way:
You have $\frac{n^2(n+1)^2}{4}+(n+1)^3$ and you want $\frac{(n+1)^2(n+2)^2}{4}$
So manipulate it to get there;
$\frac{n^2(n+1)^2}{4}+(n+1)^3 =$
$\frac{(n^2 + 4n + 4)(n+1)^2 - (4n + 4)(n+1)^2}{4}+(n+1)^3 =$
$\frac{(n+2)^2(n+1)^2}{4}- \frac{ (4n + 4)(n+1)^2}{4}+(n+1)^3 =$
$\frac{(n+2)^2(n+1)^2}{4}-(n + 1)(n+1)^2+(n+1)^3 =$
$\frac{(n+2)^2(n+1)^2}{4}-(n+1)^3+(n+1)^3 =$
$\frac{(n+2)^2(n+1)^2}{4}$
