Question

Given, $\dfrac{9}{{1!}} + \dfrac{{19}}{{2!}} + \dfrac{{35}}{{3!}} + \dfrac{{57}}{{4!}} + .....\infty$ is
A.$12e$
B.$12e - 4$
C.$12e - 3$
D.$12e - 5$

Answer 1

There are various concepts used to solve this problem. We should have the idea of factorial, expansion of exponential and about AP series. We are going to find out the ${n^{th}}$ term of the given series. After finding the ${n^{th}}$term, we have to apply the summation and try to solve the expression using the factorial concept. After these steps we will get your answer.

Complete Step by step answer:
The given series is $\dfrac{9}{{1!}} + \dfrac{{19}}{{2!}} + \dfrac{{35}}{{3!}} + \dfrac{{57}}{{4!}} + .....\infty$.
Firstly, let us try to find the ${n^{th}}$term of the series. For this consider all the numerators as
S = 9 + 19 + 35 + 57 + .....{t_n}$$$$.........(1)
We can also write equation $(1)$ as
S = {\text{ }}9 + 19 + 35 + 57 + .....{t_n}$$$$.......(2)
Subtracting equation $(1)$ and $(2)$, we will get
$0 = 9 + 10 + 16 + 22 + 28 + .... + {t_n}$
Or we can write the above expression as,
${t_n} = 9 + 10 + 16 + 22 + 28 + .... + {t_{n - 1}}$
From this expression we can observe that difference between the two consecutive numbers is $6$ except for the first two terms i.e.$9$ and $10$ . If we leave $9$ as it is then rest of the series will form an AP series where the common difference is $6$.
Sum of the AP series $= \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$
So applying the formula for sum of AP series, we will get
$\Rightarrow {t_n} = 9 + \dfrac{{n - 1}}{2}\left[ {2 \times 10 + (n - 1 - 1)6} \right]$
From the series we can observe that first term i.e.$a = 10$ and the common difference i.e. $d = 6$ . Substituting these values we will get
$\Rightarrow {t_n} = 9 + \dfrac{{n - 1}}{2}\left[ {20 + (n - 2)6} \right]$
${t_n} = 9 + \dfrac{{n - 1}}{2}\left[ {20 + (n - 2)6} \right]$
Taking $2$ common from the brackets, we get
$\Rightarrow {t_n} = 9 + (n - 1)\left[ {10 + (n - 2)3} \right]$
$\Rightarrow {t_n} = 9 + (n - 1)\left[ {10 + 3n - 6} \right]$
$\Rightarrow {t_n} = 9 + (n - 1)\left[ {3n + 4} \right]$
Simplifying the above expression, we will get
$\Rightarrow {t_n} = 9 + 3{n^2} + 4n - 3n - 4$
$\Rightarrow {t_n} = 3{n^2} + n + 5$
It is clear that ${t_n}$ is the ${n^{th}}$ term of the numerator, and from the series it is sure that ${n^{th}}$ term of the denominator is $n!$ .
So we can say that ${n^{th}}$ term of the series is
$\Rightarrow {T_n} = \dfrac{{3{n^2} + n + 5}}{{n!}}$
To further solve this, we have to simplify the equation
$\Rightarrow {T_n} = \dfrac{{3{n^2}}}{{n!}} + \dfrac{n}{{n!}} + \dfrac{5}{{n!}}$
We know that the factorial of any number is solved as
$n! = 1 \times 2 \times 3 \times ......n$
Applying the same, we will get
$\Rightarrow {T_n} = \dfrac{{3n}}{{(n - 1)!}} + \dfrac{1}{{(n - 1)!}} + \dfrac{5}{{n!}}$
Now add and subtract $\dfrac{3}{{(n - 1)!}}$ to simplify the above equation, we will get
$\Rightarrow {T_n} = \dfrac{{3n}}{{(n - 1)!}} + \dfrac{1}{{(n - 1)!}} + \dfrac{5}{{n!}} + \dfrac{3}{{(n - 1)!}} - \dfrac{3}{{(n - 1)!}}$
Simplifying the terms, we get
$\Rightarrow {T_n} = \dfrac{{3n - 3}}{{(n - 1)!}} + \dfrac{4}{{(n - 1)!}} + \dfrac{5}{{n!}}$
$\Rightarrow {T_n} = \dfrac{3}{{(n - 2)!}} + \dfrac{4}{{(n - 1)!}} + \dfrac{5}{{n!}}$
Sum of the series will be $= \sum {{T_n}}$
$\Rightarrow \sum {\dfrac{3}{{(n - 2)!}} + } \sum {\dfrac{4}{{(n - 1)!}} + \sum {\dfrac{5}{{n!}}} }$
After putting the values of$n$, we will get
$\Rightarrow 3(1 + \dfrac{1}{1} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ....) + 4(1 + \dfrac{1}{1} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ....) + 5(\dfrac{1}{1} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ....)$
And we know that the expansion of $e$ is $= (1 + \dfrac{1}{1} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ....)$
Substitute this $e$ value in the above, we get
$\Rightarrow 3(1 + \dfrac{1}{1} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ....) + 4(1 + \dfrac{1}{1} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ....) + 5(\dfrac{1}{1} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ....)$
$= 3e + 4e + 5(e - 1)$
$= 12e - 5$
Hence we can conclude that option $(D)$ is correct.

Note:
Factorial is a very useful concept. Mostly it is used in permutation and combination. It is used to identity that in how many ways we can arrange the items. Let us say we have $5$ different objects, then we can arrange them in way like for the first we have $5$ items, for second we have $4$ and so on, i.e. $5!$