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The sum of 3/1^3+5/(1^3+2^3)+7/(1^3+2^3+...

The sum of `3/1^3+5/(1^3+2^3)+7/(1^3+2^3+3^3)+ ......` upto `11` terms is equal to

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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