202009290114 Exercise 2.3.1

Prove that

$1^2+2^2+\cdots +n^2=\displaystyle{\frac{n(n+1)(2n+1)}{6}}$

for $n\in\mathbb{N}$ .

Proof.

Let $P(n)$ be the statement

$1^2+2^2+\cdots +n^2=\displaystyle{\frac{n(n+1)(2n+1)}{6}}$ for $n\in\mathbb{N}$ .

When $n=1$ , check for the validity of $P(1)$ :

$\begin{aligned} \textrm{LHS} & = 1^2 = 1 \\ \textrm{RHS} & = \frac{(1)\big( (1)+1\big) \big( 2(1)+1\big) }{6} = 1 \\ \end{aligned}$

$\because \textrm{LHS}=\textrm{RHS}$

$\therefore P(n)$ is true for $n=1$ .

Suppose $P(n)$ holds true, let’s see if $P(n+1)$ holds too:

$P(n+1):\qquad 1^2+2^2+\cdots + n^2 + (n+1)^2 \stackrel{?}{=} \displaystyle{\frac{(n+1)\big( (n+1) +1 \big) \big( 2(n+1) +1 \big) }{6}}$

Beginning with the left hand side,

$\begin{aligned} \textrm{LHS} & = 1^2+2^2+\cdots + n^2 + (n+1)^2 \\ & = P(n) + (n+1)^2 \\ & = \frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\ & = \frac{n(n+1)(2n+1)+6(n+1)^2}{6} \\ & = \frac{(n+1)\big[ (n)(2n+1) + 6(n+1) \big]}{6} \\ & = \frac{(n+1)(2n^2+n+6n+6)}{6} \\ & = \frac{(n+1)(2n^2+7n+6)}{6} \\ & = \frac{(n+1)(2n+3)(n+2)}{6} \\ \end{aligned}$