Confused on an algebraic step of inductive proof

Question

The part I've boxed is what is throwing me - why is it 6(k+1) instead of 6(k+1)^2?

Answer 1

Let's add some steps in: \begin{align*} 1^2 + 2^2 + \cdots + k^2 + (k + 1)^2 &= \frac{k(k+1)(2k+1)}{6} + (k+1)^2 \\ &= \frac{(k+1)}{6} \cdot\frac{k(2k+1)}{1} + (k+1)(k+1) \\ &= \frac{(k+1)}{6}k(2k+1) + \frac{6(k+1)}{6}(k+1) \\ &= \frac{(k+1)}{6}[k(2k+1) + 6(k+1)]. \end{align*} It's not squared, because a factor of $(k + 1)$ needed to be factorised out, leaving only one of the factors of $(k + 1)$ behind.