Question

The sum of first 9 terms of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots \dots$ is

• A. 71
• B. 96
• C. 142
• D. 192

Answer 1

The correct option is B 96

Given series is $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots \infty$
Let $T_n$ be the nth term of the given series.
$\therefore T_n=\frac{1^3+2^3+3^3+\dots \dots +n^3}{1+3+5+\dots \dots +uptonterms}$
$=\frac{{\left\{\frac{n(n+1)}{2}\right\}}^2}{n^2}=\frac{(n+1{)}^2}{4}$
$S_9={\sum }_{n=1}^9\frac{(n+1{)}^2}{4}=\frac{1}{4}(2^2+3^3+\dots \dots +{10}^2)+\frac{1^2}{4}-\frac{1^2}{4}$
$=\frac{1}{4}[\frac{10(10+1)(20+1)}{6}-1]=\frac{384}{4}=96$