Question

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

• A. 142
• B. 192
• C. 71
• D. 96

Answer 1

The correct option is D 96

$n^{th}$ term of series $=\frac{{\left[\frac{n(n+1)}{2}\right]}^2}{n^2}=\frac{1}{4}(n+1{)}^2$
Sum of n term $=\sum \frac{1}{4}(n+1{)}^2=\frac{1}{4}[\sum n^2+2\sum n+1]=\frac{1}{4}[\frac{n(n+1)(2n+1)}{6}+\frac{2n(n+1)}{2}+n]$
Sum of 9 terms
$=\frac{1}{4}[\frac{9\times 10\times 19}{6}+\frac{18\times 10}{2}+9]=\frac{384}{4}=96$