Question

$1+5+14+30+...n$term $=$

• A. $\frac{(n+2)(n+3)}{12}$
• B. $\frac{n(n+1)(n+5)}{12}$
• C. $\frac{n(n+2)(n+3)}{12}$
• D. $\frac{n{(n+1)}^2(n+2)}{12}$

Answer 1

The correct option is D

$\frac{n{(n+1)}^2(n+2)}{12}$

Explanation for the correct option:

Step 1. Find its general term of given series

Let $S_n=1+5+14+30+...n$

$=1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+........n$

Hence, $r^{th}$ term is given by $T_r=1^2+2^2+3^3+.......+r^2$

$=\frac{r\left(r+1\right)\left(2r+1\right)}{6}$

Step 2. Find the expression sum $S_n$:

$S_n=\sum _{r=1}^n\left(T_r\right)=\sum _{r=1}^n\left(\frac{r\left(r+1\right)\left(2r+1\right)}{6}\right)$

$=\frac{1}{6}\sum _{r=1}^n\left(r^2+r\right)\left(2r+1\right)$

$=\frac{1}{6}\sum _{r=1}^n\left(2r^3+3r^2+r\right)$

Step 3. Put the standard values of $\underset{}{\overset{}{\sum r^3}},\underset{}{\overset{}{\sum r^2}},\sum _{}^{}r$ and simplify

$S_n=\frac{1}{6}\left[\underset{r=1}{\overset{n}{2\sum }}\left(r^3\right)+\sum _{r=1}^n\left(r^2\right)+\underset{r=1}{\overset{n}{\sum \left(r\right)}}\right]$

$=\frac{1}{6}\left[2{\left(\frac{n(n+1)}{2}\right)}^2+3\left(\frac{n(n+1)(2n+1)}{6}\right)+\left(\frac{n(n+1)}{2}\right)\right]$

. $=\frac{1}{6}\left[\frac{{\left(n(n+1)\right)}^2}{2}+\frac{n(n+1)(2n+1)}{2}+\frac{n(n+1)}{2}\right]$

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

. $=\frac{n(n+1)}{12}\left(n^2+n+2n+2\right)$

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

. $=\frac{n(n+1)(n+2)(n+1)}{12}$

. $=\frac{\mathbf{n}{\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}^{\mathbf{2}}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}{\mathbf{12}}$

Hence, option ‘D’ is Correct.