Question & Answer
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) \[\dfrac{1}{6}n\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)\]

ANSWER Verified Verified
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
\[\begin{align}
  & \sum\limits_{r=1}^{n}{r}=\dfrac{n(n+1)}{2} \\
 & \sum\limits_{r=1}^{n}{{{r}^{2}}}=\dfrac{n(n+1)(2n+1)}{6} \\
 & \sum\limits_{r=1}^{n}{{{r}^{3}}}={{\left[ \dfrac{n(n+1)}{2} \right]}^{2}} \\
\end{align}\]

Complete step-by-step answer:
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
\[\begin{align}
  & \sum\limits_{r=1}^{n}{r}=\dfrac{n(n+1)}{2} \\
 & \sum\limits_{r=1}^{n}{{{r}^{2}}}=\dfrac{n(n+1)(2n+1)}{6} \\
 & \sum\limits_{r=1}^{n}{{{r}^{3}}}={{\left[ \dfrac{n(n+1)}{2} \right]}^{2}} \\
\end{align}\]