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

Explanation

We have to find sum of the series `(1^3)/1+(1^3+2^3)/(1+3)+(1^3+2^3+3^3)/(1+3+5)+`up to `n` terms

The given series can be written as

`(1^3+2^3=3^3+ ldots +n^3)/(1+3+5+ ldots +(2n-1))=([(n(n+1))/2]^2)/(1+3+5+ ldots (2n-1))`

In this the denominator is in A.P. series

`implies 1+3+5+ ldots +(2n-1)=n/2[2xx1+(n-1)2]=n^2`

Then

`a_n=(n^2(n+1)^2)/(4n^2)=((n+1)^2)/4=1/4 n^2+1/2 n+1/4`

`S_n=sum_(r=1)^(n)(1/4 r^2+1/2 r+1/4)`

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