Proof By Induction: Summation of Polynomial Prove by induction (weak or strong) that:  

$$\sum_{k=0}^{n-1}(k + 1)^2= \frac{n(n + 1)(2n+1)}{6}$$

My base case is:  

$n = 1$, which is true.

In my Inductive Step, I assume that: $$S(n)=\frac{n(n + 1)(2n+1)}{6}$$ holds for an arbitrary value of $n$.  

Proving it then holds for $n+1$:
$$ S(n+1)=\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$$
$$ \phantom{S(n+1)}=\frac{(n+1)((n+2)(2n+2+1)}{6}$$
$$ \phantom{S(n+1)}=\frac{(n+1)((n+2)(2n+3)}{6}$$
$$ \phantom{S(n+1)}=\frac{2n^3+9n^2+13n+6}{6}$$

but can't see how my definition of $S(n)$ can be substituted into this final equation?

[EDIT] This isn’t a duplicate because the original summation of $(k+1)^2$ is what I’m originally provided with. The apparent duplicate question also doesn’t have a correct proof by induction answer associated with it.
Hint:
\begin{align}
S(n+1)&amp;=S(n)+(n+1)^2=\frac{n(n+1)(2n+1)}6+(n+1)^2\\
&amp;=\frac{n+1}6\bigl(n(2n+1)+6(n+1)\bigr)=\frac{(n+1)(2n^2+7n+6)}6.
\end{align}
Now observe that $n=-2$ is a root of the second factor, so it is divisible by $n+2$.
Hint:
\begin{align}
S(n+1)&amp;=S(n)+(n+1)^2=\frac{n(n+1)(2n+1)}6+(n+1)^2\\
&amp;=\frac{n+1}6\bigl(n(2n+1)+6(n+1)\bigr)=\frac{(n+1)(2n^2+7n+6)}6.
\end{align}
Now observe that $n=-2$ is a root of the second factor, so it is divisible by $n+2$.