Find the sum to `n` terms of the series: `3/(1^2 .2^2)+5/(2^2 .3^2)+7/(3^2 .4^2)+`

Explanation

Given,`3/(1^2 .2^2)+5/(2^2 .3^2)+7/(3^2 .4^2)+`

=`3/(1^2 .2^2)+5/(2^2 .3^2)+7/(3^2 .4^2)+...((2n+1)/(n^2.(n+1)^2))`

now`3/(1^2 2^2)=(2^2-1^2)/(1^2 2^2)=(1/1^2​)−(1/2^2​)`

similarly,`5/(2^2 .3^2)=(1/1^2​)−(1/3^2​)`

∴sum=`(1/1^2​)−(1/2^2​)+(1/2^2​)−(1/3^2​))+.......+[(1/n^2)−(1/(n+1)^2​)]`
=`1−(1/(n+1)^2​)=((n^2+2n​)/(n+1)^2)`