Question

For all positive integral values of n, the value of $3.1.2+3.2.3+3.3.4+...+3.n.(n+1)$ is

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

Answer 1

The correct option is A

$n(n+1)(n+2)$

Explanation for correct option.

Step1. Finding ${\mathit{n}}^{\mathbf{t}\mathbf{h}}$ term

Given, series $3.1.2+3.2.3+3.3.4+...+3.n.(n+1)$

Therefore general term is $T_r=3\times r\times (r+1)$

Step2. Finding sum the series

Sum $=\sum _{r=1}^n{T}_r$

$\begin{array}{rcl}S& =& \sum _{r=1}^n{T}_r\\ & =& \underset{r=1}{\overset{n}{3\sum }}(r^2+r)\\ & =& 3\{\sum _{r=1}^n{r}^2+\sum _{r=1}^{n}r\}\\ & =& 3\{\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\}\\ & =& 3n(n+1)\{\frac{2n+1}{6}+\frac{1}{2}\}\\ & =& \frac{3n(n+1)\{2n+4\}}{6}\\ & =& n(n+1)(n+2)\end{array}$

Hence, correct option is $\mathbf{\left(}\mathit{A}\mathbf{\right)}$