Question

$\frac{1^3}{1}+\frac{(1^3+2^3)}{2}+\frac{(1^3+2^3+3^3)}{3}+....uptonterms=$

• A. $\frac{n(n+1)(n+2)(5n+3)}{48}$
• B. $\frac{n(n+1)(n+2)(n+3)}{24}$
• C. $\frac{n(n+1)(n+2)(7n+1)}{48}$
• D. $\frac{n(n+1)(n+2)(3n+5)}{48}$

Answer 1

The correct option is D

$\frac{n(n+1)(n+2)(3n+5)}{48}$

Explanation for the correct option:

Step 1. Find the $nth$term of the given series:

Let $S_n=\frac{1^3}{1}+\frac{(1^3+2^3)}{2}+\frac{(1^3+2^3+3^3)}{3}+....$ up to $n$ terms

$n^{th}$ term of given series is $T_n=\frac{1^3+2^3+3^3+\dots n^3}{n}$

$$\begin{gathered} =\frac{{\left({\displaystyle \frac{n(n+1)}{2}}\right)}^2}{n} \\ =\frac{n{(n+1)}^2}{4} \end{gathered}$$ $\left[\mathbf{\because }\mathbf{\sum }_{}{\mathit{n}}^{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{\left(}\frac{\mathbf{n}\left(n+1\right)}{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}}\right]$

$$\begin{gathered} =\frac{n(n^2+2n+1)}{4} \\ =\frac{(n^3+2n^2+n)}{4} \end{gathered}$$

Step 2. Find the sum of the series:

$S_n=\sum T_n$

$=\sum \left[\frac{(n^3+2n^2+n)}{4}\right]$

Using formulae,

$=\left[\frac{{\left({\displaystyle \frac{n(n+1)}{2}}\right)}^2}{4}+\frac{{\displaystyle \frac{2\left[n(n+1)(2n+1)\right]}{6}}}{4}+\frac{{\displaystyle \frac{n(n+1)}{2}}}{4}\right]$ $\left[\sum _{}{n}^3={\left(\frac{n\left(n+1\right)}{2}\right)}^2;\sum _{}{n}^2=\frac{n(n+1)(2n+1)}{6};\sum _{}n=\frac{n(n+1)}{2}\right]$

$=\left[\frac{n(n+1)}{2}\right]\left[\left(\frac{n(n+1)}{4}\right)+\left(\frac{4(2n+1)}{3}\right)+\left(\frac{1}{2}\right)\right]$

$=\left(\frac{n(n+1)}{4}\right)\left(\frac{3n^2+11n+10}{12}\right)$

$=\left(\frac{n(n+1)}{4}\right)\left(\frac{3n^2+5n+6n+10}{12}\right)$

$=\left(\frac{n(n+1)}{4}\right)\left[\frac{n(3n+5)+2(3n+5)}{12}\right]$

$=\frac{\mathit{n}\mathbf{(}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathit{n}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{3}\mathit{n}\mathbf{+}\mathbf{5}\mathbf{)}}{\mathbf{48}}$

Hence, Option ‘D’ is Correct.