summation - Induction Proof: $sum_{k=1}^n k^2$

Prove by induction, the following:

$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}6$$ So this is what I have so far:

We will prove the base case for $n=1$:

$$\sum_{k=1}^1 1^2 = \frac{1(1+1)(2(1)+1)}6$$ We can see this is true because $1=1$.

Using induction we can assume the statement is true for $n$, we want to prove the statement holds for the case $n+1$:

$$\begin{align*} \sum_{k=1}^{n+1} k^2 :& =\sum_{k=1}^n k^2 + (n+1)^2\\ & = \frac{n(n+1)(2n+1)}6 + (n+1)^2 && \text{by I.H}\\ & = \frac{(n+6)(2n+1)(n+1)^3}6 && \text{algebra} \end{align*}$$

This is as far as I have gotten in the proof I know that I am almost there, but I am wondering if I missed a step or did the algebra wrong. I know that what I'm supposed to get is $\frac{(n+1)(n+2)(2n+3)}6$ or is that incorrect too? This should be fairly simple, but for some reason it just isn't working for me.

Answer

Without seeing details of your "algebra" step, it is hard to be more specific. Start with:

$$\begin{align} \frac{n(n+1)(2n+1)}{6}+(n+1)^2 &= \frac{n(n+1)(2n+1)}{6} + \frac{6(n+1)^2}{6} \\ &= \frac{n+1}{6}(n(2n+1) +6(n+1)) \end{align}$$