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} +...$ is

• A. 71
• B. 96
• C. 142
• D. 192

Answer 1

The Correct Option is B

PLAN write the nth term of the given series and simplify it to get its $ \hspace15mm$ lowest form. Then, apply, $S_n = \sum T_n$ Given series is $ \frac{1^3}{1} + \frac{1^3+2^3}{1+3} + \frac{1^3+2^3+3^3}{1+3+5} +...\infty $ Let $T_n$ be the nth term of the given series. $\therefore \, \, T_n =\frac{1^3+2^3+3^3+...+n^3}{1+3+5+...+\, upto\, n\, terms}$ $\hspace10mm =\frac{\bigg\{\frac{n\, (n+1)}{2}\bigg\}^2}{n^2} = \frac{(n+1)^2}{4}$ $ S_9= \displaystyle \sum^9_{n = 1} \frac{(n+1)^4}{4} =\frac{1}{4} (2^2 +3^2 +...+10^2)+1^2-1^2]$ $ \hspace5mm = \frac{1}{4} \bigg[\frac{10(10+1)(20+1)}{6}-1\bigg]=\frac{384}{4}=96$