The sum of first 9 terms of the series `(1^3)/1+(1^3+2^3)/(1+3)+(1^3+2^3+3^3)/(1+2+3)+. . . . .` is (1) 71 (2) 96 (3) 142 (4) 192

Explanation

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

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

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

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

`sum_(x=1)^9 T_n = sum_(n=1)^9 (n+1)^2/4`

`= 1/4 sum_(n=1)^9 (n+1)^2`

= `1/4 ( 2^2 + 3^2 + 4^2 + ...+ 10^2)`

`= 1/4 ( 1^2 + 2^2 + 3^2 + .... + 10^2 - 1^2)`

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