How do you find the partial sum of $Sigma 2n$ from n=1 to 100?

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Answer 1 · 2018-02-20T00:14:19

$sum_(n=1)^100 \ 2n =10100$

We seek the sum:

$S_100 = sum_(n=1)^100 \ 2n$

We can readily find the general sum:

$S_n = sum_(r=1)^n \ 2r$
$\ \ \ \ = 2 \ sum_(r=1)^n \ r$

And. we use the standard formula for the sum of the first $n$ integers:

$sum_(r=1)^n \ r = 1/2n(n+1)$

Giving us:

$S_n = 2 * 1/2n(n+1)$
$\ \ \ \ = n(n+1)$

So with $n=100$ we have:

$S_100 = 100(101)$
$\ \ \ \ \ \ \ = 10100$

How do you find the partial sum of Sigma 2nSigma 2n from n=1 to 100?