Part b of a partial fractions Q

[Sonal7]

I have done part a) but please could you a clue on part b)
I got the partial fractions are 1/(2r+1) -1/(2r+3)

 

[MarkFL]

Okay, using the result of part a), we may write:

[MATH]\frac{3}{(2r+1)(2r+3)}=\frac{3}{2}\left(\frac{2}{(2r+1)(2r+3)}\right)=\frac{3}{2}\left(\frac{1}{2r+1}-\frac{1}{2r+3}\right)[/MATH]
And so our sum becomes:

[MATH]S_n=\frac{3}{2}\sum_{r=1}^{n}\left(\frac{1}{2r+1}-\frac{1}{2r+3}\right)=\frac{3}{2}\left(\sum_{r=1}^{n}\left(\frac{1}{2r+1}\right)-\sum_{r=1}^{n}\left(\frac{1}{2r+3}\right)\right)[/MATH]
Suppose we re-index the second sum as follows:

[MATH]S_n=\frac{3}{2}\left(\sum_{r=1}^{n}\left(\frac{1}{2r+1}\right)-\sum_{r=2}^{n+1}\left(\frac{1}{2(r-1)+3}\right)\right)[/MATH]
Can you proceed from there, stripping the first term from the first sum and the last term from the second so that you get two sums that add to zero, and you are left with just the two terms you stripped off?

 

[Sonal7]

I very appreciated your extensive help on this, let me see if this works out as I think it might be that the middle terms cancel out.
Will work out the solution now.

 

[Sonal7]

Yes I got this now to be n/(2n+3). I used method of differences. The terms in the middle cancelled out. Many thanks for the clue.

 

[MarkFL]

Let me check:

[MATH]S_n=\frac{3}{2}\left(\frac{1}{3}+\sum_{r=2}^{n}\left(\frac{1}{2r+1}\right)-\sum_{r=2}^{n}\left(\frac{1}{2r+1}\right)-\frac{1}{2n+3}\right)[/MATH]
[MATH]S_n=\frac{3}{2}\left(\frac{1}{3}-\frac{1}{2n+3}\right)[/MATH]
[MATH]S_n=\frac{3}{2}\left(\frac{2n}{3(2n+3)}\right)=\frac{n}{2n+3}\quad\checkmark[/MATH]
Good work!

 

[Sonal7]

Thanks to you! Yes it feels good. Have a good day/evening.

 

[Sonal7]

Let me check:

[MATH]S_n=\frac{3}{2}\left(\frac{1}{3}+\sum_{r=2}^{n}\left(\frac{1}{2r+1}\right)-\sum_{r=2}^{n}\left(\frac{1}{2r+1}\right)-\frac{1}{2n+3}\right)[/MATH]
[MATH]S_n=\frac{3}{2}\left(\frac{1}{3}-\frac{1}{2n+3}\right)[/MATH]
[MATH]S_n=\frac{3}{2}\left(\frac{2n}{3(2n+3)}\right)=\frac{n}{2n+3}\quad\checkmark[/MATH]
Good work!

This is so clever