# How do you find the partial sum of Sigma (2n+5) from n=1 to 20?

Mar 23, 2017

$\left(\frac{20}{2}\right) \left(7 + 45\right) = 520$

#### Explanation:

The partial sum formula for n terms looks like this:
${\sum}_{n} = \frac{n}{2} \left({a}_{1} + {a}_{n}\right)$ In words, it's the number of terms times the average of the first and last term.

In your case, with n=20, you need to find what $2 n + 5$ is for $n = 1$ and $n = 20$.
When $n = 1 , 2 n + 5 = 7.$ Likewise, when $n = 20 , 2 n + 5 = 45.$

Your formula ends up looking like this: $\left(\frac{20}{2}\right) \left(7 + 45\right) = 520$.

Mar 23, 2017

${\sum}_{1}^{20} \left(2 n + 5\right) = 520$

#### Explanation:

Before we proceed a few identities

${\sum}_{1}^{n} {n}^{0} = {\sum}_{1}^{n} 1 = n$

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

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

${\sum}_{1}^{n} {n}^{3} = {\left(\frac{n \left(n + 1\right)}{2}\right)}^{2}$

${\sum}_{1}^{n} {n}^{4} = \frac{n \left(n + 1\right) \left(2 n + 1\right) \left(3 {n}^{2} + 3 n - 1\right)}{6}$

Hence, ${\sum}_{1}^{n} \left(2 n + 5\right)$

= $2 {\sum}_{1}^{n} n + 5 \times {\Sigma}_{1}^{n} 1$

= $2 \times \frac{n \left(n + 1\right)}{2} + 5 n$

= ${n}^{2} + n + 5 n$

= ${n}^{2} + 6 n$

and ${\sum}_{1}^{20} \left(2 n + 5\right) = {20}^{2} + 6 \times 20 = 520$