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

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Answer 1 · 2017-02-18T19:10:09

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

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

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:

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

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