How do i solve the following summation problem?

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1 Answer
Truong-Son N.
May 9, 2018

There are few enough terms to just do it.

11∑n=4(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⋅122=66, so

4+...+11=1+...+11−(1+2+3)

=66−1−2−3=60.

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

11∑n=4(30−4n)=240−240=0

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