 QUESTION

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

Hint: First, we will write the ${{n}^{th}}$ term for the series and then use the below given formula to evaluate the summation.
The formula for calculating such a sum, we can use the summation which is as follows
$\sum\limits_{r=1}^{n}{{{n}^{th}}\ term}$
Some standard summations are as follows
\begin{align} & \sum\limits_{r=1}^{n}{r}=\dfrac{n(n+1)}{2} \\ & \sum\limits_{r=1}^{n}{{{r}^{2}}}=\dfrac{n(n+1)(2n+1)}{6} \\ & \sum\limits_{r=1}^{n}{{{r}^{3}}}={{\left[ \dfrac{n(n+1)}{2} \right]}^{2}} \\ \end{align}

As mentioned in the question, we have to find the summation of the given series.
So, firstly, we will find the ${{n}^{th}}$ term of the given equation as follows
${{n}^{th}}~term=\left( r \right)\cdot \left( r+1 \right)\cdot \left( r+2 \right)$
Now, using the summation formula as mentioned in the hint as follows
\begin{align} & =\sum\limits_{r=1}^{n}{\left( r \right)\cdot \left( r+1 \right)\cdot \left( r+2 \right)} \\ & =\sum\limits_{r=1}^{n}{\left( {{r}^{2}}+r \right)\cdot \left( r+2 \right)} \\ & =\sum\limits_{r=1}^{n}{\left( {{r}^{3}}+3{{r}^{2}}+2r \right)} \\ \end{align}
Using the standard summation values as mentioned in the hint, we can write the above summation as follows
\begin{align} & =\sum\limits_{r=1}^{n}{\left( {{r}^{3}}+3{{r}^{2}}+2r \right)} \\ & ={{\left( \dfrac{n(n+1)}{2} \right)}^{2}}+3\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+2\dfrac{n(n+1)}{2} \\ & =\dfrac{{{\left( n(n+1) \right)}^{2}}}{4}+\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{2}+n(n+1) \\ & =n(n+1)\left[ \dfrac{n(n+1)}{4}+\dfrac{2n+1}{2}+1 \right] \\ & =n(n+1)\left[ \dfrac{{{n}^{2}}+n}{4}+n+\dfrac{3}{2} \right] \\ & =n(n+1)\left[ \dfrac{{{n}^{2}}+5n+6}{4} \right] \\ & =\dfrac{1}{4}n(n+1)\left( n+2 \right)\left( n+3 \right) \\ \end{align}
Hence, this is the solution to the summation.

Note: The students can make an error if they don’t know the formula for the summation of a series or the standard solutions to summations that are given in the hint as follows
The formula for calculating such a sum, we can use the summation which is as follows
$\sum\limits_{r=1}^{n}{{{n}^{th}}\ term}$
Some standard summations are as follows
\begin{align} & \sum\limits_{r=1}^{n}{r}=\dfrac{n(n+1)}{2} \\ & \sum\limits_{r=1}^{n}{{{r}^{2}}}=\dfrac{n(n+1)(2n+1)}{6} \\ & \sum\limits_{r=1}^{n}{{{r}^{3}}}={{\left[ \dfrac{n(n+1)}{2} \right]}^{2}} \\ \end{align}