Question

 Find the sum of the series \[\dfrac{3}{{1!}} + \dfrac{5}{{2!}} + \dfrac{9}{{3!}} + \dfrac{{15}}{{4!}} + \dfrac{{23}}{{5!}} + ... + \infty \].
A) \[4e - 1\]
B) \[4e - 3\]
C) \[3e + 2\]
D) \[3e + 4\]

Answer 1

Hint:
In order to solve the question, first check whether the numerators form an arithmetic progression. Next, assume the \[{n}^{th}\] term of the AP. Now, consider the first three terms and find the value of the \[{t_n}\]. Finally, find the required sum of the series.

Formula Used:
\[e{\rm{ }} = {\rm{ }}1 + \dfrac{1}{{1!}}{\rm{ }} + \dfrac{1}{{2!}}{\rm{ }} + \dfrac{1}{{3!}}{\rm{ }} + \dfrac{1}{{4!}}{\rm{ }} + \ldots \]
\[e-1 = \dfrac{1}{{n!}}{\rm{ }} \]
\[2e = \dfrac{n^2}{{n!}}{\rm{ }} \]


Complete step-by-step answer:
Given that
\[\dfrac{3}{{1!}} + \dfrac{5}{{2!}} + \dfrac{9}{{3!}} + \dfrac{{15}}{{4!}} + \dfrac{{23}}{{5!}} + ... + \infty \]
Here the numerators are in AP since they have a successive difference.
Suppose that the \[{n}^{th}\] term is
\[{t_n} = a{n}^2 + bn + c\]
Let, \[n = 1,2,3\]
That is
\[{t_1} = a + b + c\]
\[ \Rightarrow a + b + c = 3\]. . . . . . (1)
 \[{t_2} = 4a + 2b + c\]
\[ \Rightarrow 4a + 2b + c = 5\]. . . . . . (2)
 \[{t_3} = 9a + 3b + c\]
\[ \Rightarrow 9a + 3b + c = 9\]. . . . . . (3)
Solving the equations (1), (2) and (3) we get
\[a = 1, b = - 1, c = 3\]
That is
\[{t_n} = {n^2} - n + 3\]
So we can write it as,
\[\dfrac{3}{{1!}} + \dfrac{5}{{2!}} + \dfrac{9}{{3!}} + \dfrac{{15}}{{4!}} + \dfrac{{23}}{{5!}} + ... + \infty \]
\[ = \dfrac{{{t_n}}}{{n!}}\]
\[ = \dfrac{{{n^2} - n + 3}}{{n!}}\]
\[ = \sum\limits_{n = 1}^\infty {\dfrac{{{n^2}}}{{n!}}} - \sum\limits_{n = 1}^\infty {\dfrac{n}{{n!}}} + 3\sum\limits_{n = 1}^\infty {\dfrac{1}{{n!}}} \]
\[ = \left( {\dfrac{1}{{1!}} + \dfrac{{{2^2}}}{{2!}} + \dfrac{{{3^2}}}{{3!}} + ... + \infty } \right) - \left( {\dfrac{1}{{1!}} + \dfrac{2}{{2!}} + \dfrac{3}{{3!}} + ... + \infty } \right) + 3\left( {\dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ... + \infty } \right)\]
\[ = 2e{\rm{ }}-{\rm{ }}e{\rm{ }} + {\rm{ }}3\left( {e - 1} \right)\]
\[ = 4e - 3\]
Hence option B is the correct answer.



Note:
Students can make mistakes while finding the value of the \[{n}^{th}\] term. Proper care should be given while solving the equations.