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.
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