How to prove using math induction that $\forall n\in \mathbb{N}$, $\sum ^{n}_{i=1}i^{2}=\frac{1}{6}n\left( n+1\right) \left(2n +1\right)$?

Question

Use mathematical induction to prove that $\forall n\in \mathbb{N}$, $$\sum ^{n}_{i=1}i^{2}=\dfrac {n\left( n+1\right) \left(2n +1\right) }{6}$$

Answer 1

True for $n=1$. Then, assuming the induction hypothesis (IH) holds for $n \geq 1$, $$\begin{align*} \sum_{k=1}^{n+1} k^2 &= \sum_{k=1}^{n} k^2 + (n+1)^2 \stackrel{\rm(IH)}{=} \frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\ &= \frac{n+1}{6}\left(n(2n+1)+6(n+1)\right) = \frac{n+1}{6}\left(2n^2+7n+6\right) \end{align*}$$ and you conclude by observing that $(n+2)(2(n+1)+1) = 2n^2+7n+6$ indeed, then by invoking the induction principle.