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