How do you use sigma notation to write the sum for #1/(13)+1/(24)+1/(35)+...+1/(1012)#?

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Answer 1 · 2017-02-14T05:43:55

#sum_(n=1)^12 1/(n(n+2))#

#sum_(n=1)^12 1/(n(n+2))#

Answer 2 · 2017-02-14T06:28:18

I give the sum here. #1/2(1+1/2 -1/11-1/12)=175/264#

The #sum# notation has already appeared, in the other answer.

As a matter of interest, I perform summation.

#1/(n(n+2))=1/2(1/n-1/(n+2))#.

So, the sum is

#1/2(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/9-1/11+1/10-1/12)#

#=1/2(1+1/2 -1/11-1/12)=175/264#

How do you use sigma notation to write the sum for #1/(1*3)+1/(2*4)+1/(3*5)+...+1/(10*12)#?