Is it true that ...? #1

Let $S_n = 1^2 + 2^2 + 3^2 + \cdots + n^2$ , We can prove the claim that $S_n = \dfrac {n(n+1)(2n+1)}6$ is true by induction .

First note that $S_n = 1^2 = 1 = \dfrac {1(2)(3)}6$ . Therefore, the claim is true for $n=1$ . Now, assuming that the claim is true for $n$ , then

$\begin{aligned} S_{\color{#D61F06}n+1} & = S_n + (n+1)^2 \\ & = \frac {n(n+1)(2n+1)}6 + (n+1)^2 \\ & = \frac {(n+1)(2n^2+n+6n+6)}6 \\ & = \frac {(n+1)(2n^2+7n+6)}6 \\ & = \frac {(n+1)(n+2)(2n+3)}6 \\ & = \frac {{\color{#D61F06}(n+1)}({\color{#D61F06}(n+1)}+1)(2{\color{#D61F06}(n+1)}+1)}6 \end{aligned}$

The claim is also true for $n+1$ , therefore it is true for all $n \ge 1$ .