Question

Find the sum of the below infinite series.
\[\dfrac{{\left( {{\text{ }}1{\text{ }} \times {\text{ }}2{\text{ }}} \right)}}{{{\text{ }}1!{\text{ }}}} + {\text{ }}\dfrac{{\left( {{\text{ }}2{\text{ }} \times {\text{ }}3{\text{ }}} \right)}}{{2!}} + {\text{ }}\dfrac{{\left( {{\text{ }}3{\text{ }} \times {\text{ }}4{\text{ }}} \right)}}{{3!}} + ....{\text{ }}\infty \]
\[\left( 1 \right)\] $ e $
\[\left( 2 \right)\] $ 2e $
\[\left( 3 \right)\] $ 3e $
\[\left( 4 \right)\]\[none{\text{ }}of{\text{ }}these\]

Answer 1

Hint: We have to find the sum of the given infinite series . We solve this question using the concept of sum to $ n $ terms of special series . We would simplify the given equation to form an AP and on solving and equating the equations we can find the sum of infinite terms of series , using first term \[\left( a \right)\] and common difference \[\left( d \right)\]. We will put the value of $ a $ and $ d $ in the formula of sum of infinite sum of series.

Complete step-by-step answer:
Given series : is \[\dfrac{{\left( {{\text{ }}1{\text{ }} \times {\text{ }}2{\text{ }}} \right)}}{{{\text{ }}1!{\text{ }}}} + {\text{ }}\dfrac{{\left( {{\text{ }}2{\text{ }} \times {\text{ }}3{\text{ }}} \right)}}{{2!}} + {\text{ }}\dfrac{{\left( {{\text{ }}3{\text{ }} \times {\text{ }}4{\text{ }}} \right)}}{{3!}} + ....{\text{ }}\infty \]
Here the denominator terms are \[1!,{\text{ }}2!,{\text{ }}3!, \ldots \]
So ,
The denominator of $ {n^{(th)}}term = n! $
Now ,
Consider the terms in the numerator \[1{\text{ }} \times {\text{ }}2{\text{ }},{\text{ }}2{\text{ }} \times {\text{ }}3{\text{ }},{\text{ }}3{\text{ }} \times {\text{ }}4{\text{ }}, \ldots \ldots \ldots \ldots \ldots \ldots \ldots \]
Here ,
The general formula for the numerator becomes\[n{\text{ }} \times {\text{ }}\left( {{\text{ }}n{\text{ }} + {\text{ }}1{\text{ }}} \right)\], where n is a natural number
So , the $ {n^{(th)}}term $ of the series would be
\[{T_n}{\text{ }} = {\text{ }}\dfrac{{n{\text{ }} \times {\text{ }}\left( {{\text{ }}n{\text{ }} + {\text{ }}1{\text{ }}} \right){\text{ }}}}{{\left( {{\text{ }}n!{\text{ }}} \right)}}\]Expanding the term of factorial
\[n{\text{ }}!{\text{ }} = {\text{ }}n{\text{ }} \times {\text{ }}\left( {n - 1} \right){\text{ }} \times {\text{ }}\left( {n - 2} \right){\text{ }} \times {\text{ }} \ldots \ldots \ldots \ldots \ldots .{\text{ }} \times {\text{ }}3{\text{ }} \times {\text{ }}2{\text{ }} \times {\text{ }}1\]
We get the $ {n^{(th)}}term $ of the series as ,
\[{T_n}{\text{ }} = {\text{ }}\dfrac{{{\text{ }}\left( {{\text{ }}n{\text{ }} + {\text{ }}1{\text{ }}} \right)}}{{\left( {{\text{ }}n{\text{ }} - {\text{ }}1} \right)!}}\]
Put various values of $ n $ for different terms
Put \[n{\text{ }} = {\text{ }}1{\text{ }},{\text{ }}2{\text{ }},{\text{ }}3{\text{ }},{\text{ }}4{\text{ }},{\text{ }} \ldots \ldots \ldots \ldots \ldots \ldots \]
Putting values in \[{T_n}\] and representing it in terms such that it becomes factorial of exponential series
For , \[n{\text{ }} = {\text{ }}1\]
\[{T_1}{\text{ }} = {\text{ }}\dfrac{{\left( {{\text{ }}1{\text{ }} + {\text{ }}1{\text{ }}} \right)}}{{0!}}\]
\[{T_1}{\text{ }} = {\text{ }}\dfrac{2}{{0!}}\]
For , \[n{\text{ }} = {\text{ }}2\]
\[{T_2}{\text{ }} = {\text{ }}\dfrac{3}{{1!}}\]
\[{T_2}{\text{ }} = {\text{ }}\dfrac{1}{{0!}}{\text{ }} + {\text{ }}\dfrac{2}{{1!}}\]
For , \[n{\text{ }} = {\text{ }}3\]
\[{T_3}{\text{ }} = {\text{ }}\dfrac{4}{{2!}}\]
\[{T_3}{\text{ }} = {\text{ }}\dfrac{1}{{1!}} + {\text{ }}\dfrac{2}{{2!}}\]
For , \[n{\text{ }} = {\text{ }}4\]
\[{T_4}{\text{ }} = {\text{ }}\dfrac{5}{{3!}}\]
\[{T_4}{\text{ }} = {\text{ }}\dfrac{1}{{2!}}{\text{ }} + {\text{ }}\dfrac{2}{{3!}}\]
Sum of the series \[ = {\text{ }}{T_1}{\text{ }} + {\text{ }}{T_2} + {\text{ }}{T_3}{\text{ }} + {\text{ }} \ldots \ldots \ldots ..{\text{ }}\infty \]
Sum of series \[ = {\text{ }}\left( {{\text{ }}\dfrac{1}{{0!}}{\text{ }} + {\text{ }}\dfrac{1}{{1!}}{\text{ }} + \dfrac{1}{{2!}} + \ldots {\text{ }}\infty } \right){\text{ }} + {\text{ }}\left( {{\text{ }}\dfrac{2}{{0!}}{\text{ }} + {\text{ }}\dfrac{2}{{1!}} + {\text{ }}\dfrac{2}{{2!}}{\text{ }} + \ldots \infty } \right)\]
Taking $ 2 $ common , we get
Sum of series $ = \left( {{\text{ }}\dfrac{1}{{0!}}{\text{ }} + {\text{ }}\dfrac{1}{{1!}}{\text{ }} + \dfrac{1}{{2!}} + \ldots {\text{ }}\infty } \right){\text{ }} + 2 \times \left( {{\text{ }}\dfrac{1}{{0!}}{\text{ }} + {\text{ }}\dfrac{1}{{1!}}{\text{ }} + \dfrac{1}{{2!}} + \ldots {\text{ }}\infty } \right){\text{ }} $
As , the factorial expansion of exponential function :
\[e{\text{ }} = {\text{ }}\left( {{\text{ }}1{\text{ }} + {\text{ }}1{\text{ }} + {\text{ }}\dfrac{1}{{2!}}{\text{ }} + {\text{ }}\dfrac{1}{{3!}}{\text{ }} + {\text{ }}\dfrac{1}{{4!}} + \dfrac{1}{{5!}}{\text{ }} + \ldots .{\text{ }}} \right)\]
Using this , we get
Sum of series \[ = {\text{ }}e{\text{ }} + {\text{ }}2{\text{ }}e\]
Sum of series \[ = {\text{ }}3{\text{ }}e\]
Hence , the sum of the given series is \[3{\text{ }}e\].
Thus , the correct option is \[\left( 3 \right)\]
So, the correct answer is “Option 3”.

Note: The expansion of exponential series :
\[e{\text{ }} = {\text{ }}\dfrac{1}{{n!}}{\text{ }} = {\text{ }}\dfrac{1}{1}{\text{ }} + {\text{ }}\dfrac{1}{1}{\text{ }} + {\text{ }}\dfrac{1}{2}{\text{ }} + {\text{ }}\dfrac{1}{6}{\text{ }} + {\text{ }}...\]
 $ {e^{( - 1)}} = \dfrac{{{{( - 1)}^n}}}{{n!}} = \dfrac{1}{1} - \dfrac{1}{1} + \dfrac{1}{2} - \dfrac{1}{6} + ... $
\[{e^x} = \dfrac{{{x^n}}}{{n!}} = \dfrac{1}{1} + \dfrac{x}{1} + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{6} + ...\]
We used the concept of $ {n^{(th)}}term $ of AP , the formula of sum of n terms of an AP, the concept of special series .
The sum of first $ n $ terms of some special series :
\[1{\text{ }} + {\text{ }}2{\text{ }} + {\text{ }}3{\text{ }} + {\text{ }}4{\text{ }} + {\text{ }} \ldots \ldots \ldots \ldots \ldots .{\text{ }} + {\text{ }}n{\text{ }} = {\text{ }}\dfrac{{n\left( {n + 1} \right)}}{2}\]
 $ {1^2} + {2^2} + {3^3} + ........ + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6} $
 $ {1^3} + {2^3} + {3^3} + ............. + {n^3} = \dfrac{{{{[n(n + 1)]}^2}}}{4} $