Question

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

Answer 1

`sum_(n=0)^50 (1000-5n) = 44625`

Explanation:

`sum_(n=0)^50 (1000-5n) = sum_(n=0)^50 (1000) - 5sum_(n=0)^50(n) =`

We can use the standard result `sum_(i=1)^n i = 1/2n(n+1)`, and note that:

`sum_(n=0)^50 (1000) = 1000+1000+ ... +1000 \ \` (51 terms)
`sum_(i=0)^n i=0+ sum_(i=1)^n`

So we get:

`sum_(n=0)^50 (1000-5n) = (51)(1000) - 5*1/2(50)(50+1)`
`" "= 51000 - 5/2(50)(51)`
`" "= 51000 - 6375`
`" " = 44625`