<s>[INST] Induction to prove that $1^2+ 3^2 + 5^2+ \cdots +(2n + 1)^2=\frac{(n+1)(2n+1)(2n + 3)}{3}$ Have I started this right? I know I have to add $(k+1)$ but why?

Use mathematical induction to prove that $$1^2+ 3^2 + 5^2+ \cdots +(2n + 1)^2=\frac{(n+1)(2n+1)(2n + 3)}{3}.$$

BASIS STEP: $P(0)$ is true since $$\begin{align}(2(0)+1)^2 &= \frac{(0+1)(2(0)+1)(2(0)+3)}{3}\\1 &= \frac{3}{3}\\1 &= 1.\end{align}$$
INDUCTIVE STEP: Assume $P(k)$ holds for an arbitrary integer $k>0$. $$1^2+ 3^2 + 5^2+ \cdots +(2k + 1)^2=\frac{(k+1)(2k+1)(2k + 3)}{3}$$
 [/INST] You've proved that the statement is true for $k=0$ and now supposing that is true to an arbitrary $k$, 
$$1^2+2^2+\cdots+(2k+1)^2 = \frac{(k+1)(2k+1)(2k+3)}{3}$$
let's see the case $k+1$. Then we use the hypothesis of induction and we get
$$\begin{align}1^2+2^2+\cdots+(2k+1)^2+(2(k+1)+1)^2 &= \frac{(k+1)(2k+1)(2k+3)}{3}+(2(k+1)+1)^2\\ &= \frac{(k+1)(2k+1)(2k+3)+3(2k+3)^2}{3}\\ &= \frac{[(k+1)(2k+1)+3(2k+3)](2k+3)}{3}\\ &= \frac{[2k^2+3k+1+6k+9](2k+3)}{3}\\ &= \frac{(2k+5)(k+2)(2k+3)}{3}\\ &= \frac{[(k+1)+1][2(k+1)+1][2(k+1)+3]}{3},\end{align}$$
where we used that $(2k+5)(k+2)=2k^2+9k+10.$
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