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

Question

I'm asked to prove $$\sum_{k=1}^{n} k^{2} = \frac{1}{6} n(n+1)(2n+1)$$ using proof by induction.

Now, I know how to do induction proofs and I end up at this step, needing to prove that:

$$\frac{1}{6} n(n+1)(2n+1) + (n+1)^2 = \frac{1}{6} (n+1)(n+2)(2n+3)$$

So $$n(2n+1) + n+1 = (n+2)(2n+3)$$ $$2n^2 + 2n + 1 = 2n^2 + 7n + 6$$

which is obviously not equal.

What am I doing wrong?

Answer 1

Dividing by $(n+1)$ was good, but when you got rid of the $\frac16$ factor, you forgot to multiply the $(n+1)^2$ term by $6$.

Answer 2

The issue is that the $(n+1)^2$ term on the left does not have a factor of $\frac16$.

Answer 3

There is a simple visual proof of this: link