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 :
$1{\text{ }} + {\text{ }}2{\text{ }} + {\text{ }}3{\text{ }} + {\text{ }}4{\text{ }} + {\text{ }} \ldots \ldots \ldots \ldots \ldots .{\text{ }} + {\text{ }}n{\text{ }} = {\text{ }}\dfrac{{n\left( {n + 1} \right)}}{2}$
${1^2} + {2^2} + {3^3} + ........ + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}$
${1^3} + {2^3} + {3^3} + ............. + {n^3} = \dfrac{{{{[n(n + 1)]}^2}}}{4}$