# How do you find the sum of the series n^2 from n=1 to 6?

Mar 5, 2017

$91$

#### Explanation:

Adding the individual terms in the sum.

${\sum}_{n = 1}^{6} {n}^{2}$

$= {1}^{2} + {2}^{2} + {3}^{2} + {4}^{2} + {5}^{2} + {6}^{2}$

$= 91$

$\textcolor{b l u e}{\text{OR}}$

$\textcolor{red}{\text{Knowledge of the summation identity}}$

${\sum}_{n = 1}^{6} {n}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$

Using n = 6

$= \frac{1}{6} \times 6 \left(6 + 1\right) \left(12 + 1\right) = 1 \times 7 \times 13$

$= 91$