# How do you use the properties of summation to evaluate the sum of Sigma (i^2+3) from i=1 to 20?

Nov 10, 2016

${\sum}_{i = 1}^{20} \left({i}^{2} + 3\right) = 2930$

#### Explanation:

You can use the standard result

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

So:

${\sum}_{i = 1}^{20} \left({i}^{2} + 3\right) = {\sum}_{i = 1}^{20} \left({i}^{2}\right) + {\sum}_{i = 1}^{20} \left(3\right)$
$\therefore {\sum}_{i = 1}^{20} \left({i}^{2} + 3\right) = \frac{1}{6} \left(20\right) \left(21\right) \left(41\right) + \left(20\right) \left(3\right)$
$\therefore {\sum}_{i = 1}^{20} \left({i}^{2} + 3\right) = 2870 + 60$
$\therefore {\sum}_{i = 1}^{20} \left({i}^{2} + 3\right) = 2930$