Question

$\frac{1^3}{1}+\frac{1^3+2^3}{2}+\frac{1^3+2^3+3^3}{3}+...n$ terms $=$

• A. $\frac{n(n+1)(n+2)(3n+5)}{48}$
• B. $\frac{n(n+2)(n+3)(4n+5)}{48}$
• C. $\frac{n(n+3)(n+5)(3n+7)}{48}$
• D. $\frac{n(n+2)(n+3)(3n+5)}{48}$

Answer 1

The correct option is A $\frac{n(n+1)(n+2)(3n+5)}{48}$

$T_n=\frac{\sum n^3}{n}=\frac{n(n+1{)}^2}{4}=\frac{1}{4}(n^3+2n^2+n)$
Therefore,
Required summation is $=\sum _{k=1}^n{T}_k=\frac{1}{4}[\sum k^3+2\sum k^2+\sum k]$
$=\frac{1}{4}[\frac{n^2(n+1{)}^2}{4}+2\cdot \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}]$
$=\frac{1}{4}n(n+1)[\frac{n(n+1)}{4}+\frac{(2n+1)}{3}+\frac{1}{2}]$
$=\frac{n(n+1)}{48}\left[3n\right(n+1)+4(2n+1)+6]$
$=\frac{n(n+1)}{48}(3n^2+11n+10)$
$=\frac{n(n+1)(n+2)(3n+5)}{48}$