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

Hint: 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.