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Question: The coefficient of \({x^3}\) in the expansion of \({e^{2x + 3}}\) as a series in powers of x is A....

The coefficient of x3{x^3} in the expansion of e2x+3{e^{2x + 3}} as a series in powers of x is
A. e36\dfrac{{{e^3}}}{6}
B. 2e36\dfrac{{2{e^3}}}{6}
C. 4e36\dfrac{{4{e^3}}}{6}
D. 8e36\dfrac{{8{e^3}}}{6}

Explanation

Solution

At first we’ll write the expansion of any function using the Maclaurin’s series which is f(x)=f(0)+f′(0)1!x+f′′(0)2!x2+f′′′(0)3!x3+......+fn(0)n!xnf(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 x3{x^3}. Then substituting the given function i.e. e2x+3{e^{2x + 3}} in place of f(x) we’ll find the coefficient of x3{x^3}in the expansion of e2x+3{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)+f′(0)1!x+f′′(0)2!x2+f′′′(0)3!x3+......+fn(0)n!xnf(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 e2x+3{e^{2x + 3}}
Let f(x)=e2x+3f(x) = {e^{2x + 3}}
Therefore, on differentiating with-respect-to x , we get,
⇒f′(x)=2e2x+3\Rightarrow f'(x) = 2{e^{2x + 3}}
Again on differentiating with-respect-to x, we get,
⇒f′′(x)=4e2x+3\Rightarrow f''(x) = 4{e^{2x + 3}}
Again on differentiating with-respect-to x, we get,
⇒f′′′(x)=8e2x+3\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
x3{x^3}is given by f′′′(0)3!\dfrac{{f'''(0)}}{{3!}}
Therefore the coefficient of x3{x^3} in the expansion of e2x+3=8e2(0)+33!{e^{2x + 3}} = \dfrac{{8{e^{2(0) + 3}}}}{{3!}}
=8e36= \dfrac{{8{e^3}}}{6}
Hence, The coefficient of x3{x^3} in the expansion of e2x+3{e^{2x + 3}} as a series in powers of x is
8e36\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)+f′(0)1!x+f′′(0)2!x2+f′′′(0)3!x3+......+fn(0)n!xnf(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 e2x+3{e^{2x + 3}}
Let f(x)=e2x+3f(x) = {e^{2x + 3}}
Therefore, on differentiating with-respect-to x
⇒f′(x)=2e2x+3\Rightarrow f'(x) = 2{e^{2x + 3}}
Again on differentiating with-respect-to x
⇒f′′(x)=22e2x+3\Rightarrow f''(x) = {2^2}{e^{2x + 3}}
Again on differentiating with-respect-to x
⇒f′′′(x)=23e2x+3\Rightarrow f'''(x) = {2^3}{e^{2x + 3}}
Therefore we can say that fn(x)=2ne2x+3{f^n}(x) = {2^n}{e^{2x + 3}}
Substituting all these values in the Maclaurin’s series
⇒e2x+3=e2(0)+3+2e2(0)+31!x+22e2(0)+32!x2+23e2(0)+33!x3+......+2ne2(0)+3n!xn\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}
⇒e2x+3=e3+2e31!x+22e32!x2+23e33!x3+......+2ne3n!xn\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 ofe2x+3=23e33!{e^{2x + 3}} = \dfrac{{{2^3}{e^3}}}{{3!}}
=8e36= \dfrac{{8{e^3}}}{6}
Therefore, option(D) is correct