Question

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

Answer 1

Hint: At first we’ll write the expansion of any function using the Maclaurin’s series which is $f(x) = f(0) + \dfrac{{f'(0)}}{{1!}}x + \dfrac{{f''(0)}}{{2!}}{x^2} + \dfrac{{f'''(0)}}{{3!}}{x^3} + ...... + \dfrac{{{f^n}(0)}}{{n!}}{x^n}$, in the series, we’ll find the coefficient of ${x^3}$. Then substituting the given function i.e. ${e^{2x + 3}}$ in place of f(x) we’ll find the coefficient of ${x^3}$in the expansion of ${e^{2x + 3}}$.

Complete step by step Answer:

We know that according to Maclaurin’s series expansion of any function is given by
$f(x) = f(0) + \dfrac{{f'(0)}}{{1!}}x + \dfrac{{f''(0)}}{{2!}}{x^2} + \dfrac{{f'''(0)}}{{3!}}{x^3} + ...... + \dfrac{{{f^n}(0)}}{{n!}}{x^n}$
Now we have the function ${e^{2x + 3}}$
Let $f(x) = {e^{2x + 3}}$
Therefore, on differentiating with-respect-to x , we get,
$ \Rightarrow f'(x) = 2{e^{2x + 3}}$
Again on differentiating with-respect-to x, we get,
$ \Rightarrow f''(x) = 4{e^{2x + 3}}$
Again on differentiating with-respect-to x, we get,
$ \Rightarrow f'''(x) = 8{e^{2x + 3}}$
Now, according to the Maclaurin’s series expansion of any function, we can say that the coefficient of
${x^3}$is given by $\dfrac{{f'''(0)}}{{3!}}$
Therefore the coefficient of ${x^3}$ in the expansion of ${e^{2x + 3}} = \dfrac{{8{e^{2(0) + 3}}}}{{3!}}$
$ = \dfrac{{8{e^3}}}{6}$
Hence, The coefficient of ${x^3}$ in the expansion of ${e^{2x + 3}}$ as a series in powers of x is
$\dfrac{{8{e^3}}}{6}$
Therefore, option(D) is correct.

Note: An alternative way to find the answer can be given as follows:
We know that according to Maclaurin’s series expansion of any function is given by
$f(x) = f(0) + \dfrac{{f'(0)}}{{1!}}x + \dfrac{{f''(0)}}{{2!}}{x^2} + \dfrac{{f'''(0)}}{{3!}}{x^3} + ...... + \dfrac{{{f^n}(0)}}{{n!}}{x^n}$

Now we have the function ${e^{2x + 3}}$
Let $f(x) = {e^{2x + 3}}$
Therefore, on differentiating with-respect-to x
$ \Rightarrow f'(x) = 2{e^{2x + 3}}$
Again on differentiating with-respect-to x
$ \Rightarrow f''(x) = {2^2}{e^{2x + 3}}$
Again on differentiating with-respect-to x
$ \Rightarrow f'''(x) = {2^3}{e^{2x + 3}}$
Therefore we can say that ${f^n}(x) = {2^n}{e^{2x + 3}}$
Substituting all these values in the Maclaurin’s series
$ \Rightarrow {e^{2x + 3}} = {e^{2(0) + 3}} + \dfrac{{2{e^{2(0) + 3}}}}{{1!}}x + \dfrac{{{2^2}{e^{2(0) + 3}}}}{{2!}}{x^2} + \dfrac{{{2^3}{e^{2(0) + 3}}}}{{3!}}{x^3} + ...... + \dfrac{{{2^n}{e^{2(0) + 3}}}}{{n!}}{x^n}$
$ \Rightarrow {e^{2x + 3}} = {e^3} + \dfrac{{2{e^3}}}{{1!}}x + \dfrac{{{2^2}{e^3}}}{{2!}}{x^2} + \dfrac{{{2^3}{e^3}}}{{3!}}{x^3} + ...... + \dfrac{{{2^n}{e^3}}}{{n!}}{x^n}$
Therefore from the expansion of${e^{2x + 3}} = \dfrac{{{2^3}{e^3}}}{{3!}}$
$ = \dfrac{{8{e^3}}}{6}$
Therefore, option(D) is correct