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The sum of the series 3/1^2+5/(1^2+2^2)+...

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

Text Solution

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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)`
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