How do you evaluate the series $Sigma 5r$ from r=3 to 8?

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Answer 1 · 2017-01-03T22:54:12

$sum_(r=3)^8 5r = 165$

We seek:

$sum_(r=3)^8 5r = 5sum_(r=3)^8 r$

Due to the small number of terms required we can just expand the individual terms to get:

$sum_(r=3)^8 5r = 5(3+4+5+6+7+8}$
$" " = 5(33)$
$" " = 165$

If the number of individual terms were larger this would be quite cumbersome and use of the standard summation formula

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

would be more appropriate. If we used this approach we would get

$sum_(r=3)^8 5r = 5{(sum_(r=1)^8) r - (sum_(r=1)^2 r )}$
$" " = 5{1/2*8*9 - 1/2*2*3}$
$" " = 5/2{72 - 6}$
$" " = 5/2*66$
$" " = 165$

How do you evaluate the series Sigma 5rSigma 5r from r=3 to 8?