The sum of the 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. 71
B. 96
C. 142
D. 192

Answer

B

Explanation

`T_(n)=(1^3)+2^(3)+3^(3)+….+n^(3))/(1+3+5+…+(2n-1))=((n^(2)(n+1)^(2))/4)/(n/2(1+2n-1))`
`rArrT_(n)=((n^(2)(n+1)^(2))/4)/(n^(2))`
`thereforeT_(n)=1/4(n+1)^(2)`
`T_(n)=1/4[n^(2)+2n+1]`
`S_(n)=sum_(n=1)^(n)T_(n)`
`S_(n)=1/4[(n(n+1)(2n+1))/6+n(n+1)+n]`
Putting n=9, we get
`S_(9)=1/4[(9xx10xx19)/6+9xx10+9]`
`=1/4[285+90+9]=384/4=96`