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+3+5)+"........"` is

Options

A. 192
B. 71
C. 96
D. 142

Answer

C

Explanation

`T_n = (1^2 +2^2 +3^3 +...+n^3)/(1+3+5+...+(2n-1))=((n(n+1))/2)^2/((n(1+2n-1))/2)=((n+1)^2)/4`
` 1/4 (n^2+2n+1)`
`:." "S_n = 1/4(Sigman^2 +2Sigman+Sigma1)=1/4[(n(n+1)(2n+1))/6+(2n(n+1))/2+n]`
`S_9 = 1/4 [285 +90 +9] =96`