Question

The coefficient of ${x^n}$ in the expansion of $\dfrac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$ is

Answer 1

At first we will simplify the given expression as the sum of two terms.
Then we’ll write the expansion of any function using the Maclaurin’s series, in the series, we’ll find the coefficient of ${x^n}$, using this we will find the coefficient of ${x^n}$ in both the terms we had on simplifying the given expression.
Now, the sum of those coefficients is the coefficient of ${x^n}$ in the given expression and hence we will get the answer.

Complete step-by-step answer:
Given: The expression $\dfrac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$
Now solving for $\dfrac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$
On splitting the fraction we get,
$\Rightarrow \dfrac{{{e^{7x}}}}{{{e^{3x}}}} + \dfrac{{{e^x}}}{{{e^{3x}}}}$
We know that $\dfrac{{{x^a}}}{{{x^b}}} = {x^{a - b}}$ , using this we get,
$\Rightarrow {e^{4x}} + {e^{ - 2x}}$
Now, we know that according to Maclaurin’s series expansion of any function is given by
$f(x) = f(0) + \dfrac{{f'(0)}}{{1!}}x + \dfrac{{f''(0)}}{{2!}}{x^2} + \dfrac{{f'''(0)}}{{3!}}{x^3} + ...... + \dfrac{{{f^n}(0)}}{{n!}}{x^n}$
For the first term of the expression i.e. ${e^{4x}}$
Let $f(x) = {e^{4x}}$
Therefore, on differentiating with-respect-to x
$\Rightarrow f'(x) = 4{e^{4x}}$
Again on differentiating with-respect-to x
$\Rightarrow f''(x) = {4^2}{e^{4x}}$
Again on differentiating with-respect-to x
$\Rightarrow f'''(x) = {4^3}{e^{4x}}$
From the above derivatives, we can say that
$\Rightarrow {f^n}(x) = {4^n}{e^{4x}}$
Now, according to the Maclaurin’s series expansion of any function, we can say that the coefficient of ${x^n}$ is given by $\dfrac{{{f^n}(0)}}{{n!}}$
Therefore the coefficient of ${x^n}$ in the expansion of ${e^{4x}} = \dfrac{{{4^n}{e^{4(0)}}}}{{n!}}$
$= \dfrac{{{4^n}}}{{n!}}$
Now, or the first term of the expression i.e. ${e^{ - 2x}}$
Let $g(x) = {e^{ - 2x}}$
Therefore, on differentiating with-respect-to x
$\Rightarrow g'(x) = - 2{e^{ - 2x}}$
Again on differentiating with-respect-to x
$\Rightarrow g''(x) = {\left( { - 2x} \right)^2}{e^{ - 2x}}$
Again on differentiating with-respect-to x
$\Rightarrow g'''(x) = {\left( { - 2x} \right)^3}{e^{ - 2x}}$
From the above derivatives, we can say that
$\Rightarrow {g^n}(x) = {\left( { - 2x} \right)^n}{e^{ - 2x}}$
Now, according to the Maclaurin’s series expansion of any function, we can say that the coefficient of ${x^n}$ is given by $\dfrac{{{g^n}(0)}}{{n!}}$
Therefore the coefficient of ${x^n}$ in the expansion of ${e^{ - 2x}} = \dfrac{{{{\left( { - 2x} \right)}^n}{e^{ - 2(0)}}}}{{n!}}$
$= \dfrac{{{{\left( { - 2} \right)}^n}}}{{n!}}$
Therefore the coefficient of ${x^n}$ in the expansion of $\dfrac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$ will be $\dfrac{{{4^n}}}{{n!}} + \dfrac{{{{\left( { - 2} \right)}^n}}}{{n!}}$
$= \dfrac{1}{{n!}}\left[ {{4^n} + {{\left( { - 2} \right)}^n}} \right]$

Note: We can find the coefficient of ${x^n}$ in the expansion ${e^{4x}} + {e^{ - 2x}}$ by taking the whole expression as a function
Let $h(x) = {e^{4x}} + {e^{ - 2x}}$
Therefore, on differentiating with-respect-to x
$\Rightarrow h'(x) = 4{e^{4x}} + \left( { - 2} \right){e^{ - 2x}}$
Again on differentiating with-respect-to x
$\Rightarrow h''(x) = {\left( 4 \right)^2}{e^{4x}} + {\left( { - 2x} \right)^2}{e^{ - 2x}}$
Again on differentiating with-respect-to x
$\Rightarrow h'''(x) = {\left( 4 \right)^3}{e^{4x}} + {\left( { - 2x} \right)^3}{e^{ - 2x}}$
From the above derivatives, we can say that
$\Rightarrow {h^n}(x) = {\left( 4 \right)^n}{e^{4x}} + {\left( { - 2x} \right)^n}{e^{ - 2x}}$
Now, according to the Maclaurin’s series expansion of any function, we can say that the coefficient of ${x^n}$is given by $\dfrac{{{h^n}(0)}}{{n!}}$
Therefore the coefficient of ${x^n}$ in the expansion of ${e^{ - 2x}} = \dfrac{{{{\left( 4 \right)}^n}{e^{4(0)}} + {{\left( { - 2x} \right)}^n}{e^{ - 2(0)}}}}{{n!}}$
$= \dfrac{{{{\left( 4 \right)}^n} + {{\left( { - 2} \right)}^n}}}{{n!}}$,
which is the similar answer as of the above solution