How do i solve the following summation problem?

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Answer 1 · 2018-05-09T20:56:04

There are few enough terms to just do it.

$sum_(n=4)^(11) (30 - 4n)$

$= (30 - 4(4)) + (30 - 4(5)) + (30 - 4(6)) + (30 - 4(7)) + (30 - 4(8)) + (30 - 4(9)) + (30 - 4(10)) + (30 - 4(11))$

From term 4 through 11 inclusive, there are $8$ values of $30$ , so:

$= 8(30) - 4(4) - . . . - 4(11)$

$= 8(30) - 4(4 + . . . + 11)$

We know that $1 + . . . + 11 = (11*12)/2 = 66$ , so

$4 + . . . + 11 = 1 + . . . + 11 - (1 + 2 + 3)$

$= 66 - 1 - 2 - 3 = 60$ .

Therefore, $4(4 + . . . + 11) = 240$ , and:

$sum_(n=4)^(11) (30 - 4n) = 240 - 240 = color(blue)(0)$