Question

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

Answer 1

`sum_(r=1)^250 (1000-r) = 218625`

Explanation:

Let us denote the sum:

`S_n = sum_(r=1)^n (1000-r)`

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 = 1/2n(n+1)`

And so we get:

`S_n = sum_(r=1)^n (1000) - sum_(r=1)^n(r)`
`\ \ \ \ = 1000n - 1/2n(n+1)`

`\ \ \ \ = n/2{2000 - (n+1)}`

`\ \ \ \ = n/2(2000-n-1)`

`\ \ \ \ = n/2(1999-n)`

Using this result, the desired sum is:

`sum_(r=1)^250 (1000-r) = S_250`
`" " = 250/2(1999-250)`
`" " = 125(1749)`
`" " = 218625`