Question

Find the sum of the series $\dfrac{{12}}{{2!}} + \dfrac{{28}}{{3!}} + \dfrac{{50}}{{4!}} + \dfrac{{78}}{{5!}} + \cdots $
A. $e$
B. $4e$
C. $3e$
D. 5e

Answer 1

Hint: First we will assume that the given equation to be ${S_n}$. Then subtract ${S_n}$ from ${S_n} - 12$ to calculate the ${n^{th}}$ term of ${S_n}$. Using the ${n^{th}}$ term we will find the term of the given series by substituting for $n$. Then by using the expanded form of ${e^x}$, we can get the desired result.

Formula Used:
The ${n^{th}}$ term of an AP is ${t_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term and $d$ is the common difference.
The sum of ${n^{th}}$ term AP ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \cdots $

Complete step by step solution:
Given series is $\dfrac{{12}}{{2!}} + \dfrac{{28}}{{3!}} + \dfrac{{50}}{{4!}} + \dfrac{{78}}{{5!}} + \cdots $
Assume that ${S_n} = 12 + 28 + 50 + 78 + \cdots + {t_n}$
So, ${S_n} - 12 = 28 + 50 + 78 + \cdots + {t_n}$
Subtract ${S_n}$ by from ${S_n} - 12$
${S_n} - 12 = 28 + 50 + 78 + \cdots + {t_n}$
          ${S_n} = 12 + 28 + 50 + 78 + \cdots + {t_n}$
   $\overline { - 12 = 16 + 22 + 28 + \cdots \left( {n - 1} \right){\rm{terms}} - {t_n}} $
Add 12 both sides
$ \Rightarrow {t_n} = 12 + \left[ {16 + 22 + 28 + \cdots \left( {n - 1} \right){\rm{times}}} \right]$ ……(1)
From the above equation we can say that $22 - 16 = 28 - 16 = 6$.
So, $16 + 22 + 28 + \cdots \left( {n - 1} \right){\rm{times}}$ is an AP series
Now applying sum of ${n^{th}}$ term AP in $16 + 22 + 28 + \cdots \left( {n - 1} \right){\rm{times}}$
$16 + 22 + 28 + \cdots \left( {n - 1} \right){\rm{times}} = \dfrac{{n - 1}}{2}\left[ {2 \cdot 16 + \left( {n - 1 - 1} \right) \cdot 6} \right]$
$ \Rightarrow 16 + 22 + 28 + \cdots \left( {n - 1} \right){\rm{times}} = \dfrac{{n - 1}}{2}\left[ {32 + 6n - 12} \right]$
$ \Rightarrow 16 + 22 + 28 + \cdots \left( {n - 1} \right){\rm{times}} = \dfrac{{n - 1}}{2}\left[ {20 + 6n} \right]$
$ \Rightarrow 16 + 22 + 28 + \cdots \left( {n - 1} \right){\rm{times}} = \left( {n - 1} \right)\left( {10 + 3n} \right)$
$ \Rightarrow 16 + 22 + 28 + \cdots \left( {n - 1} \right){\rm{times}} = 3{n^2} - 3n + 10n - 10$
$ \Rightarrow 16 + 22 + 28 + \cdots \left( {n - 1} \right){\rm{times}} = 3{n^2} + 7n - 10$
Now putting $16 + 22 + 28 + \cdots \left( {n - 1} \right){\rm{times}} = 3{n^2} + 7n - 10$ in equation (1)
${t_n} = 12 + 3{n^2} + 7n - 10$
   $ = 3{n^2} + 7n + 2$
So the denominator of ${n^{th}}$ term of the given series is $3{n^2} + 7n + 2$ and the denominator is $\left( {n + 1} \right)!$.
Thus the ${n^{th}}$ term of the given series is $\dfrac{{3{n^2} + 7n + 2}}{{\left( {n + 1} \right)!}}$.
                                                                     $ = \dfrac{{3{n^2} + 3n - 3n + 7n + 2}}{{\left( {n + 1} \right)!}}$
                                                                     $ = \dfrac{{3{n^2} + 3n + 4n + 4 - 4 + 2}}{{\left( {n + 1} \right)!}}$
                                                                     $ = \dfrac{{3n\left( {n + 1} \right) + 4\left( {n + 1} \right) - 2}}{{\left( {n + 1} \right)!}}$
                                                                  $ = \dfrac{{3n\left( {n + 1} \right)}}{{\left( {n + 1} \right) \cdot n \cdot \left( {n - 1} \right)!}} + \dfrac{{4\left( {n + 1} \right)}}{{\left( {n + 1} \right) \cdot n!}} - \dfrac{2}{{\left( {n + 1} \right)!}}$
                                                                $ = \dfrac{3}{{\left( {n - 1} \right)!}} + \dfrac{4}{{n!}} - \dfrac{2}{{\left( {n + 1} \right)!}}$
Now putting $n = 1$ in the expression ${a_n} = \dfrac{3}{{\left( {n - 1} \right)!}} + \dfrac{4}{{n!}} - \dfrac{2}{{\left( {n + 1} \right)!}}$
${a_1} = 3 \cdot \dfrac{1}{{0!}} + 4 \cdot \dfrac{1}{{1!}} - 2 \cdot \dfrac{1}{{2!}}$
Now putting $n = 2$ in the expression ${a_n} = \dfrac{3}{{\left( {n - 1} \right)!}} + \dfrac{4}{{n!}} - \dfrac{2}{{\left( {n + 1} \right)!}}$
${a_2} = 3 \cdot \dfrac{1}{{1!}} + 4 \cdot \dfrac{1}{{2!}} - 2 \cdot \dfrac{1}{{3!}}$
Now putting $n = 3$ in the expression ${a_n} = \dfrac{3}{{\left( {n - 1} \right)!}} + \dfrac{4}{{n!}} - \dfrac{2}{{\left( {n + 1} \right)!}}$
${a_3} = 3 \cdot \dfrac{1}{{2!}} + 4 \cdot \dfrac{1}{{3!}} - 2 \cdot \dfrac{1}{{4!}}$
So we can rewrite the given series as
$\dfrac{{12}}{{2!}} + \dfrac{{28}}{{3!}} + \dfrac{{50}}{{4!}} + \dfrac{{78}}{{5!}} + \cdots = \left( {3 \cdot \dfrac{1}{{0!}} + 4 \cdot \dfrac{1}{{1!}} - 2 \cdot \dfrac{1}{{2!}}} \right) + \left( {3 \cdot \dfrac{1}{{1!}} + 4 \cdot \dfrac{1}{{2!}} - 2 \cdot \dfrac{1}{{3!}}} \right) + \left( {3 \cdot \dfrac{1}{{2!}} + 4 \cdot \dfrac{1}{{3!}} - 2 \cdot \dfrac{1}{{4!}}} \right) + \cdots $
$\dfrac{{12}}{{2!}} + \dfrac{{28}}{{3!}} + \dfrac{{50}}{{4!}} + \dfrac{{78}}{{5!}} + \cdots = 3 \cdot \left( {\dfrac{1}{{0!}} + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \cdots } \right) + 4 \cdot \left( {\dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + \cdots } \right) - 2 \cdot \left( {\dfrac{1}{{2!}} + \dfrac{1}{{3!}} + \dfrac{1}{{4!}} + \cdots } \right)$ …….(2)
We know that ${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \cdots $
Putting $x = 1$
$e = 1 + \dfrac{1}{{1!}} + \dfrac{{{1^2}}}{{2!}} + \dfrac{1}{{3!}} \cdots $
$ \Rightarrow e = \dfrac{1}{{0!}} + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} \cdots $
$ \Rightarrow e - 1 = \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} \cdots $ [Since $\dfrac{1}{{0!}} = 1$]
$ \Rightarrow e - 2 = \dfrac{1}{{2!}} + \dfrac{1}{{3!}} \cdots $ [Since $\dfrac{1}{{1!}} = 1$]
Now putting the values $e = \dfrac{1}{{0!}} + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} \cdots $, $e - 1 = \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} \cdots $, $e - 2 = \dfrac{1}{{2!}} + \dfrac{1}{{3!}} \cdots $ in equation (2)
$\dfrac{{12}}{{2!}} + \dfrac{{28}}{{3!}} + \dfrac{{50}}{{4!}} + \dfrac{{78}}{{5!}} + \cdots = 3 \cdot e + 4 \cdot \left( {e - 1} \right) - 2 \cdot \left( {e - 2} \right)$
$ \Rightarrow \dfrac{{12}}{{2!}} + \dfrac{{28}}{{3!}} + \dfrac{{50}}{{4!}} + \dfrac{{78}}{{5!}} + \cdots = 3e + 4e - 4 - 2e + 4$
$ \Rightarrow \dfrac{{12}}{{2!}} + \dfrac{{28}}{{3!}} + \dfrac{{50}}{{4!}} + \dfrac{{78}}{{5!}} + \cdots = 3e + 4e - 4 - 2e + 4$
$ \Rightarrow \dfrac{{12}}{{2!}} + \dfrac{{28}}{{3!}} + \dfrac{{50}}{{4!}} + \dfrac{{78}}{{5!}} + \cdots = 5e$

Option ‘D’ is correct

Note: We tend to do common mistake to substitute the value of the series and substitute $\dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} \cdots = e$ and $\dfrac{1}{{2!}} + \dfrac{1}{{3!}} \cdots = e$ which is incorrect. The correct values are $e - 1 = \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} \cdots $ and $e - 2 = \dfrac{1}{{2!}} + \dfrac{1}{{3!}} \cdots $.