Question

The coefficient of \[{x^n}\] in the series \[1 + \dfrac{{a + bx}}{{1!}} + \dfrac{{{{\left( {a + bx} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {a + bx} \right)}^3}}}{{3!}} + .....\infty \] is
A. \[\dfrac{{{{\left( {ab} \right)}^n}}}{{n!}}\]
B. \[{e^b}\dfrac{{{a^n}}}{{n!}}\]
C. \[{e^a}\dfrac{{{b^n}}}{{n!}}\]
D. \[{e^{a + b}}\dfrac{{{{\left( {a + b} \right)}^n}}}{{n!}}\]

Answer 1

- Hint: The given series is of an exponential function. So, write the series expansion for the exponential function \[{e^x}\]. Then compare this series expansion with the given series. Clearly, the series will be same if we replace \[x\] by \[\left( {a + bx} \right)\]. The \[\left( {n + 1} \right)\]-th term of the series contains \[{x^n}\]. So, the coefficient of \[{x^n}\] will be found from the \[\left( {n + 1} \right)\]-th term.

Formula Used:
The series expansion for \[{e^x}\] is \[1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + .....\infty \]

Complete step-by-step answer:
The given series is of an exponential function.
We know the series expansion for the exponential function \[{e^x}\], which is \[1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + .....\infty \]
So, \[{e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + .....\infty .........\left( i \right)\]
But there is \[\left( {a + bx} \right)\] at the place of \[x\] in the given series.
So, if we replace \[x\] by \[\left( {a + bx} \right)\], we get
\[{e^{a + bx}} = 1 + \dfrac{{a + bx}}{{1!}} + \dfrac{{{{\left( {a + bx} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {a + bx} \right)}^3}}}{{3!}} + .....\infty ..........\left( {ii} \right)\]
So, the given series is of the exponential function \[{e^{a + bx}}\]
Using the formula \[{a^{m + n}} = {a^m} \cdot {a^n}\], we can write \[{e^{a + bx}} = {e^a} \cdot {e^{bx}}\]
Replacing \[x\] by \[bx\] in the equation \[\left( i \right)\], we get
\[{e^{bx}} = 1 + \dfrac{{bx}}{{1!}} + \dfrac{{{{\left( {bx} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {bx} \right)}^3}}}{{3!}} + .....\infty .........\left( {iii} \right)\]
and \[{e^a}\] is constant.
So, \[{e^{a + bx}} = {e^a} \cdot {e^{bx}} = {e^a}\left\{ {1 + \dfrac{{bx}}{{1!}} + \dfrac{{{{\left( {bx} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {bx} \right)}^3}}}{{3!}} + .....\infty } \right\}.........\left( {iv} \right)\]
It is required to find out the coefficient of \[{x^n}\] in the series given by \[\left( {iv} \right)\]
The term \[{x^n}\] is present in \[\left( {n + 1} \right)\]-th term of the series given by \[\left( {iv} \right)\]
Clearly, the \[\left( {n + 1} \right)\]-th term is \[{e^a}\dfrac{{{{\left( {bx} \right)}^n}}}{{n!}}\] i.e. \[{e^a}\dfrac{{{b^n}}}{{n!}}{x^n}\]
So, the coefficient of \[{x^n}\] in the given series is \[{e^a}\dfrac{{{b^n}}}{{n!}}\]
Hence, option C is correct.

Note:- In the series expansion of \[{e^x}\], all the terms are positive for positive values of \[x\] but for the negative values of \[x\], the signs of the terms alter. The denominators consist of the factorial of the natural numbers.