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summation of series using method of differences

how to find sum of 1st n term of
1/(r(r+3)) using method of differences
Original post by Akhito Kanbara
how to find sum of 1st n term of
1/(r(r+3)) using method of differences


Start by splitting 1/(r(r+3)) into partial fractions, i.e. write it as A/r - B/(r+3), and find the values of the constants A and B. Then write out the terms explicitly using the partial fractions form, and you should get cancellation.
Original post by HapaxOromenon3
Start by splitting 1/(r(r+3)) into partial fractions, i.e. write it as A/r - B/(r+3), and find the values of the constants A and B. Then write out the terms explicitly using the partial fractions form, and you should get cancellation.

I got
1/3r-1/3(r+3)
but when i sub n=1,2,3 could not see how they cancel off
(edited 7 years ago)
Original post by Akhito Kanbara
I got
1/3r-1/3(r+3)
but when i sub n=1,2,3 could not see how they cancel off


The difference between the two denominators is +3 so you might need to write out a few more terms to clearly see the pattern. Try writing out the first and last four terms and see if you can spot the pattern and find which terms don't cancel.
Original post by ThatPerson
The difference between the two denominators is +3 so you might need to write out a few more terms to clearly see the pattern. Try writing out the first and last four terms and see if you can spot the pattern and find which terms don't cancel.

thanks i know how to do that already.

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