# Thread: [SOLVED] Method of differences problems

1. ## [SOLVED] Method of differences problems

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

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

My working :

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

Afrer doing the partial fraction , i got

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

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

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

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

My working :

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

Afrer doing the partial fraction , i got

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

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

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

and in that case

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

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,

$\displaystyle \sum^{n}_{r=1}\frac{1}{r^2}-\frac{1}{(r+1)^2} = \left( \frac{1}{1^2} - \frac{1}{2^2} \right) + \left( \frac{1}{2^2} - \frac{1}{3^2} \right) + \left( \frac{1}{3^2} - \frac{1}{4^2} \right) + \cdots + \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$
$\displaystyle \frac{1}{1^2} - \frac{1}{(n+1)^2}$