# Thread: [SOLVED] Method of differences problems

1. ## [SOLVED] Method of differences problems

If $f(r)=\frac{1}{r^2}$ , simpify f(r)-f(r+1) . Hence , find the sum of the first n terms of the series

$
\frac{3}{1^2\times2^2}+\frac{5}{2^2\times3^2}+\fra c{7}{3^2\times4^2}+...
$

My working :

After simplifying , i got $\frac{2r+1}{r^2(r+1)^2}$ .

Afrer doing the partial fraction , i got

$\sum^{n}_{r=1}\frac{2r+1}{r^2(r+1)^2}=\sum^{n}_{r= 1}[\frac{1}{r^2}-\frac{1}{r+1}\frac{1}{(r+1)^2}]$

I need help on finding the sums , i am not sure how to continue from where i stopped and i am not sure how to use the method of differences here . THanks for any help

2. Originally Posted by mathaddict
If $f(r)=\frac{1}{r^2}$ , simpify f(r)-f(r+1) . Hence , find the sum of the first n terms of the series

$
\frac{3}{1^2\times2^2}+\frac{5}{2^2\times3^2}+\fra c{7}{3^2\times4^2}+...
$

My working :

After simplifying , i got $\frac{2r+1}{r^2(r+1)^2}$ .

Afrer doing the partial fraction , i got

$\sum^{n}_{r=1}\frac{2r+1}{r^2(r+1)^2}=\sum^{n}_{r= 1}[\frac{1}{r^2}-\frac{1}{r+1}\frac{1}{(r+1)^2}]$

I need help on finding the sums , i am not sure how to continue from where i stopped and i am not sure how to use the method of differences here . THanks for any help
I think you have an extra term here. I think it should be

$\sum^{n}_{r=1}\frac{2r+1}{r^2(r+1)^2}=\sum^{n}_{r= 1}\frac{1}{r^2}-\frac{1}{(r+1)^2}$

and in that case

$\sum^{n}_{r=1}\frac{1}{r^2}-\frac{1}{(r+1)^2} = 1- \frac{1}{(n+1)^2}$

3. Thanks Danny , but can you elaborate more on your final step be4 getting the answer ? Did you apply the method of differences ? How did you do that ?

Thanks again,

4. Originally Posted by mathaddict
Thanks Danny , but can you elaborate more on your final step be4 getting the answer ? Did you apply the method of differences ? How did you do that ?

Thanks again,
Sure. Writing out the terms we find
$
\sum^{n}_{r=1}\frac{1}{r^2}-\frac{1}{(r+1)^2} = \left( \frac{1}{1^2} - \frac{1}{2^2} \right) + \left( \frac{1}{2^2} - \frac{1}{3^2} \right) + \left( \frac{1}{3^2} - \frac{1}{4^2} \right) + \cdots + \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$

notice all of the terms cancel except the first of the first term and the last of the last term leaving
$
\frac{1}{1^2} - \frac{1}{(n+1)^2}$