# How do you find the partial sum of Sigma (1000-5n) from n=0 to 50?

Feb 18, 2017

${\sum}_{n = 0}^{50} \left(1000 - 5 n\right) = 44625$

#### Explanation:

${\sum}_{n = 0}^{50} \left(1000 - 5 n\right) = {\sum}_{n = 0}^{50} \left(1000\right) - 5 {\sum}_{n = 0}^{50} \left(n\right) =$

We can use the standard result ${\sum}_{i = 1}^{n} i = \frac{1}{2} n \left(n + 1\right)$, and note that:

${\sum}_{n = 0}^{50} \left(1000\right) = 1000 + 1000 + \ldots + 1000 \setminus \setminus$ (51 terms)
${\sum}_{i = 0}^{n} i = 0 + {\sum}_{i = 1}^{n}$

So we get:

${\sum}_{n = 0}^{50} \left(1000 - 5 n\right) = \left(51\right) \left(1000\right) - 5 \cdot \frac{1}{2} \left(50\right) \left(50 + 1\right)$
$\text{ } = 51000 - \frac{5}{2} \left(50\right) \left(51\right)$
$\text{ } = 51000 - 6375$
$\text{ } = 44625$