Home
Class 12
MATHS
The sum to n terms of the series 3/(1^2)...

The sum to `n` terms of the series `3/(1^2)+5/(1^2+2^2)+7/(1^2+2^2+3^2)+-------` is

Text Solution

Verified by Experts

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)`
Promotional Banner

Similar Questions

Explore conceptually related problems

The sum to 50 terms of the series 3/1^2+5/(1^2+2^2)+7/(1^+2^2+3^2)+….+… is

The sum to 50 terms of the series 3/1^2+5/(1^2+2^2)+7/(1^+2^2+3^2)+….+… is

The sum to 50 terms of the series 3/1^2+5/(1^2+2^2)+7/(1^+2^2+3^2)+….+… is

Find the sum of first n terms of the series 3/1^2+5/(1^2+2^2)+7/(1^2+2^2+3^2) +.... .

The sum to n term of the series 1(1!)+2(2!)+3(3!)+

The sum to n term of the series 1(1!)+2(2!)+3(3!)+

The sum of n terms of the series (3)/(1^(2))+(5)/(1^(2)+2^(2))+ (7)/(1^(2)+2^(2)+3^(2))+.. .. .. is