Question

The coefficient of ${x^3}$ in the expansion of $\exp (2x + 3)$ as a series in powers of $x$ is
A.$\dfrac{{\exp (3)}}{6}$
B.$\dfrac{{2\exp (3)}}{6}$
C.$\dfrac{{4\exp (3)}}{6}$
D.$\dfrac{{8\exp (3)}}{6}$

Answer 1

To solve this question, first we will split the exponential power by applying concept of addition of powers i.e.
$\Rightarrow {e^{a + b}} = {e^a} \times {e^b}$
After that expand the exponential as a series in power of x by the formula given by:
$\Rightarrow {e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} +$……………
Now you can find the coefficient of ${x^3}$.

Complete step-by-step answer:
In this question we are given an expansion i.e. ${e^{(2x + 3)}}$and we have to find the coefficient of ${x^3}$ in this series.
First of all, let's split the exponential power by the power addition concept i.e. if we multiply two numbers having same base but their powers are different then it can be written as single number with same base and powers is addition of both the powers.
$\Rightarrow {e^{a + b}} = {e^a} \times {e^b}$
$\Rightarrow {e^{2x + 3}} = {e^{2x}} \times {e^3}$ …………..(1)
Expand the exponential as a series i.e. given by:
$\Rightarrow {e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} +$………
Replacing x with 2x we get,
$\Rightarrow {e^{2x}} = 1 + \dfrac{{2x}}{{1!}} + \dfrac{{{{(2x)}^2}}}{{2!}} + \dfrac{{(2{x^3})}}{{3!}} +$………..
By opening the bracket, we get,
$\Rightarrow {e^{2x}} = 1 + \dfrac{x}{{1!}} + \dfrac{{4{x^2}}}{{2!}} + \dfrac{{8{x^3}}}{{3!}} +$……………..
We can open the factorials in the denominator by the formula i.e. $n! = n.(n - 1).(n - 2).......2.1$. By putting this in exponential series we get,
$\Rightarrow {e^{2x}} = 1 + \dfrac{x}{1} + \dfrac{{4{x^2}}}{{2 \times 1}} + \dfrac{{8{x^3}}}{{3 \times 2 \times 1}} +$……………
Put this value in (1), we get,
$\Rightarrow {e^{2x + 3}} = {e^3} \times \left( {1 + \dfrac{x}{1} + \dfrac{{4{x^2}}}{{2 \times 1}} + \dfrac{{8{x^3}}}{{3 \times 2 \times 1}} + ...} \right)$
$\Rightarrow {e^{2x + 3}} = {e^3} \times \left( {1 + x + \dfrac{{4{x^2}}}{2} + \dfrac{{8{x^3}}}{6} + ...} \right)$
By opening the bracket, we get,
$\Rightarrow {e^{2x + 3}} = {e^3} + {e^3}.x + {e^3}.\dfrac{{4{x^2}}}{2} + {e^3}\dfrac{{8{x^3}}}{6} +$
Here in the series you can find the coefficient of ${x^3}$ term is $\dfrac{{8{e^3}}}{6}$.
Hence, option D is the correct answer.

Note: In this question students generally expand the exponential as a series without splitting the powers. This may lead to complexity and they can’t come to the result or they find it difficult to find the answer. So, in these types of questions first split the powers then solve the question. It will simplify your solution.
While giving the answer students mainly forget to multiply the coefficient by ${e^3}$ but ${e^3}$ is a constant so we will multiply it with coefficients of the terms in the series. Take care of these types of mistakes.