Question

Sum of $\frac{1^3}{1}+\frac{1^3+2^3}{1+2}+...$to $n$ terms is

• A. $\frac{(n+1)(n+2)}{3}$
• B. $n(n+1)(n+2)$
• C. $\frac{n(n+1)(n+2)}{6}$
• D. $\frac{n(n+1)(n+2)}{2}$

Answer 1

Answer: (C)

$S=\sum T_n=\sum \frac{1^3+2^3+...+n^3}{1+2+...+n}=\sum \frac{\sum n^3}{\sum n}$
$=\sum \frac{\frac{n^2{\left(n+1\right)}^2}{4}}{\frac{n\left(n+1\right)}{2}}=\sum \frac{n\left(n+1\right)}{2}=\frac{1}{2}\left[\sum n^2+\sum n\right]$
$=\frac{1}{2}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\right]$
$=\frac{n\left(n+1\right)}{4}\left[\frac{2n+1}{3}+1\right]=\frac{n\left(n+1\right)\left(2n+4\right)}{12}$
$=\frac{n\left(n+1\right)\left(n+2\right)}{6}$