Q.Using the principle of mathematical induction, prove that $1^2+2^2+3^2+\cdots+n^2 = \dfrac{n(n+1)(2n+1)}{6}$ for all natural numbers $n$.

The same two-step induction pattern is used here, except the extra term added at each stage is $(k+1)^2$ rather than $(k+1)$, which needs a slightly longer factoring step.

Basis step ($n=1$): LHS $=1^2=1$. RHS $=\dfrac{1(2)(3)}{6}=\dfrac{6}{6}=1$. So $P(1)$ is true.

Inductive step: assume $P(k)$: $1^2+2^2+\cdots+k^2 = \dfrac{k(k+1)(2k+1)}{6}$.

Add $(k+1)^2$ to both sides:

Factor $(k+1)$ out of the right side: $(k+1)\left[\dfrac{k(2k+1)}{6}+(k+1)\right] = (k+1)\cdot\dfrac{k(2k+1)+6(k+1)}{6}$.

Expand the numerator inside the bracket: $k(2k+1)+6(k+1) = 2k^2+k+6k+6 = 2k^2+7k+6$, which factors as $(2k+3)(k+2)$.

So the sum becomes $(k+1)\cdot\dfrac{(2k+3)(k+2)}{6} = \dfrac{(k+1)(k+2)(2k+3)}{6}$.

This is exactly the formula $\dfrac{n(n+1)(2n+1)}{6}$ with $n=k+1$ (since $n+1=k+2$ and $2n+1=2k+3$), so $P(k+1)$ is true. …

Errors while factoring $2k^2+7k+6$ into $(2k+3)(k+2)$ — it is worth expanding $(2k+3)(k+2)=2k^2+4k+3k+6=2k^2+7k+6$ back out as a check before …