The Sum of Squares Formula


Ken Levasseur
UMass Lowell Mathematical Sciences
http://faculty.uml.edu/klevasseur
$$6 \cdot (1^2+2^2+3^2+4^2) = 6\cdot \sum_{k=1}^4 k^2$$ $$\longrightarrow$$
$$\longrightarrow$$$$6 \cdot (1^2+2^2+3^2+4^2) = 4 (4+1)(2\cdot 4+1)$$
$$ 6 \cdot \sum_{k=1}^n k^2 = n(n+1)(2n+1)$$$$\longrightarrow$$$$ \sum_{k=1}^n k^2 =\frac{ n(n+1)(2n+1)}{6}$$

Reference: Cotangents, 16 October, 2014