Prove sum of $k^2$ using $k^3$

Question

So the title may be a little bit vague, but I am quite stuck with the following problem. Asked is to first prove that $(k + 1)^3 - k^3 = 3k^2 + 3k + 1$. This is not the problem however. The question now asks to prove that

$$ \sum_{k=1}^{n} k^2 = \frac{1}{6}n(n+1)(2n+1)$$ using the fact that $(k + 1)^3 - k^3 = 3k^2 + 3k + 1$. I have no idea however to start working on this. Anyone have any idea? Does this have anything to do with telescopic series?

Answer 1

Hint: take the sum for $k=1$ to $n$ of both sides of the equation $(k+1)^3 - k^3 = 3k^2 + 3k + 1$.

Answer 2

Hint: This is a model for the general method to find a recursive relation to calculate $$S_r(n)=\sum_{k=1}^n k^r$$ Generalising your giving relation with the binomial formula, you get $$(k+1)^{r+1}-k^{r+1}=\sum_{i=1}^{r+1}\binom{r+1}i k^{r+1-i}$$ and adding these relations, taking into account the telescoping sum in the left hand side, there results the relation $$\sum_{i=1}^r\binom{r+1}i S_i(n)=(n+1)^{r+1}-(n+1).$$

Answer 3

Elaborating on Patrick’s hint, $$\sum_{k=1}^n(3k^2+3k+1) \\ =3\sum_{k=1}^nk^2+3\sum_{k=1}^nk+\sum_{k=1}^n 1 \\ =3S +3\cdot\frac{n(n+1)}{2}+n$$ This equals the telescoping sum $$\sum_{k=1}^n[(k+1)^3-k^3]=(n+1)^3-1=n^3+3n^2+3n $$

Rearranging the terms gives the result $$S=\frac{n(n+1)(2n+1)}{6}$$