Toolkit 2

Sum of squares

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

Proof

Using the identity

(k+1)^3 - k^3 = 3k^2 + 3k + 1,

we have

2^3 - 1^3 = 3\cdot 1^2 + 3\cdot 1 + 1,

3^3 - 2^3 = 3\cdot 2^2 + 3\cdot 2 + 1,

\vdots

(n+1)^3 - n^3 = 3n^2 + 3n + 1.

Adding these identities, all intermediate terms on the left-hand side cancel, yielding

(n+1)^3 - 1 = 3(1^2 + 2^2 + \cdots + n^2) + 3(1 + 2 + \cdots + n) + n.

1 + 2 + \cdots + n = \frac{n(n+1)}{2}.

Hence,

\begin{aligned} 3(1^2 + 2^2 + \cdots + n^2) &= (n+1)^3 - 1 - \frac{3n(n+1)}{2} - n \\ &= \frac{n(n+1)(2n+1)}{2}. \end{aligned}

Therefore,

1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}. \quad\square