Proof By Induction: Summation of Polynomial

Question

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.

This should be a duplicate ! Note that the sum is equal to $$\sum_{k=1}^n k^2$$ — Peter, Aug 31 '18 at 22:35
You do not have $S(n+1)=\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$, which is NOT the induction hypothesis $S(n) = \cdots$. — , Aug 31 '18 at 22:38
@JohnMa not sure what you’re talking about? I do indeed have that as my induction hypothesis? — adamcasey, Sep 01 '18 at 19:06
$S(n+1)=\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$ is what you need to prove, @adamcasey — , Sep 01 '18 at 20:09

score -1 · Answer 1 · answered Aug 31 '18 at 23:44

-1

Hint: \begin{align} S(n+1)&=S(n)+(n+1)^2=\frac{n(n+1)(2n+1)}6+(n+1)^2\\ &=\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 $n+2$.

answered Aug 31 '18 at 23:44

Bernard

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Proof By Induction: Summation of Polynomial

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