# How do you evaluate the series Sigma 5r from r=3 to 8?

Jan 3, 2017

${\sum}_{r = 3}^{8} 5 r = 165$

#### Explanation:

We seek:

${\sum}_{r = 3}^{8} 5 r = 5 {\sum}_{r = 3}^{8} r$

Due to the small number of terms required we can just expand the individual terms to get:

${\sum}_{r = 3}^{8} 5 r = 5 \left(3 + 4 + 5 + 6 + 7 + 8\right\}$
$\text{ } = 5 \left(33\right)$
$\text{ } = 165$

If the number of individual terms were larger this would be quite cumbersome and use of the standard summation formula

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

would be more appropriate. If we used this approach we would get

${\sum}_{r = 3}^{8} 5 r = 5 \left\{\left({\sum}_{r = 1}^{8}\right) r - \left({\sum}_{r = 1}^{2} r\right)\right\}$
$\text{ } = 5 \left\{\frac{1}{2} \cdot 8 \cdot 9 - \frac{1}{2} \cdot 2 \cdot 3\right\}$
$\text{ } = \frac{5}{2} \left\{72 - 6\right\}$
$\text{ } = \frac{5}{2} \cdot 66$
$\text{ } = 165$