# How do you find the partial sum of Sigma (1000-n) from n=1 to 250?

Mar 26, 2018

${\sum}_{r = 1}^{250} \left(1000 - r\right) = 218625$

#### Explanation:

Let us denote the sum:

${S}_{n} = {\sum}_{r = 1}^{n} \left(1000 - r\right)$

Then we can use the standard result for the sum of the first $n$ integers (Using the formula for Arithmetic progression):

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

And so we get:

${S}_{n} = {\sum}_{r = 1}^{n} \left(1000\right) - {\sum}_{r = 1}^{n} \left(r\right)$
$\setminus \setminus \setminus \setminus = 1000 n - \frac{1}{2} n \left(n + 1\right)$

$\setminus \setminus \setminus \setminus = \frac{n}{2} \left\{2000 - \left(n + 1\right)\right\}$

$\setminus \setminus \setminus \setminus = \frac{n}{2} \left(2000 - n - 1\right)$

$\setminus \setminus \setminus \setminus = \frac{n}{2} \left(1999 - n\right)$

Using this result, the desired sum is:

${\sum}_{r = 1}^{250} \left(1000 - r\right) = {S}_{250}$
$\text{ } = \frac{250}{2} \left(1999 - 250\right)$
$\text{ } = 125 \left(1749\right)$
$\text{ } = 218625$