Question

the value of the sum $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + ......$ upto $n$ terms is
(A) $\dfrac{1}{6}{n^2}(2{n^2} + 1)$
(B) $\dfrac{1}{6}({n^2} - 1)(2n - 1)(2n + 3)$
(C) $\dfrac{1}{8}({n^2} + 1)({n^2} + 5)$
(D) $\dfrac{1}{4}n(n + 1)(n + 2)(n + 3)$

Answer 1

here, we use the standard formulas for summation of $n,{n^2}$and ${n^3}$ terms.

Complete step-by-step answer:
Given, the sum is $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + ......$
The ${n^{th}}$term of the sum is given by $n(n + 1)(n + 2)$
Then the sum is given by $\sum\limits_{r = 1}^n {r(r + 1)(r + 2)}$
$\Rightarrow \sum\limits_{r = 1}^n {({r^2} + r)(r + 2)}$
$\Rightarrow \sum\limits_{r = 1}^n {({r^3} + 3{r^2} + 2r)}$
$\Rightarrow \sum\limits_{r = 1}^n {{r^3}} + 3\sum\limits_{r = 1}^n {{r^2}} + 2\sum\limits_{r = 1}^n r$
Now, the sum can be found using the standard formulas for summation of $n,{n^2}$and ${n^3}$ terms which are given by
$\sum\limits_{r = 1}^n r = \dfrac{1}{2}n(n + 1)$
$\sum\limits_{r = 1}^n {{r^2}} = \dfrac{1}{6}n(n + 1)(2n + 1)$
$\sum\limits_{r = 1}^n {{r^3}} = \dfrac{1}{4}{n^2}{(n + 1)^2}$
$\Rightarrow$the sum $\sum\limits_{r = 1}^n {{r^3}} + 3\sum\limits_{r = 1}^n {{r^2}} + 2\sum\limits_{r = 1}^n r$ is given by
$\dfrac{1}{4}{n^2}{(n + 1)^2} + 3 \cdot \dfrac{1}{6}n(n + 1)(2n + 1) + 2 \cdot \dfrac{1}{2}n(n + 1)$
$\Rightarrow \dfrac{1}{4}{n^2}{(n + 1)^2} + \dfrac{1}{2}n(n + 1)(2n + 1) + n(n + 1)$
$\Rightarrow \dfrac{1}{4}n(n + 1)\left[ {n(n + 1) + 2n(2n + 1) + 4} \right]$
$\Rightarrow \dfrac{1}{4}n(n + 1)\left[ {{n^2} + n + 4n + 2 + 4} \right]$
$\Rightarrow \dfrac{1}{4}n(n + 1)\left( {{n^2} + 5n + 6} \right)$
$\Rightarrow \dfrac{1}{4}n(n + 1)({n^2} + 3n + 2n + 6)$
$\Rightarrow \dfrac{1}{4}n(n + 1)\left[ {n(n + 3) + 2(n + 3)} \right]$
$\Rightarrow \dfrac{1}{4}n(n + 1)(n + 2)(n + 3)$
Therefore, (D) $\dfrac{1}{4}n(n + 1)(n + 2)(n + 3)$ is the required solution.

Note: in these types of questions always try to find the ${n^{th}}$ term.