Answer

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**Hint:**In the above question you have to find the coefficient of ${x^n}$ in the expansion of $\dfrac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$. At first, you have to reduce the expanding term then by applying a simple law of exponent, the term will get reduced. Now find the coefficient of ${x^n}$ in the expansion of the new reduced term. So let us see how we can solve this problem.

**Step by step solution:**

In the given question we were asked to find the coefficient of ${x^n}$ in the expansion of $\dfrac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$. First of all, we will reduce the expression by the formula of $\dfrac{{x + y}}{z} = \dfrac{x}{z} + \dfrac{y}{z}$.

On applying the same formula on the expanding term we get

$= \dfrac{{{e^{7x}}}}{{{e^{3x}}}} + \dfrac{{{e^x}}}{{{e^{3x}}}}$

According to the law of exponent, $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$ . On applying the same law of exponent we get,

$= {e^{7x - 3x}} + {e^{x - 3x}}$

$= {e^{4x}} + {e^{ - 2x}}$

Now, we have reduced the term in $= {e^{4x}} + {e^{ - 2x}}$ . So we have to find the coefficient of ${x^n}$ in ${e^{4x}} + {e^{ - 2x}}$

$= \dfrac{{{{(4)}^n}}}{{n!}} + \dfrac{{{{( - 2)}^n}}}{{n!}}$

**Therefore, the coefficient of ${x^n}$ in the expansion of $\dfrac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$ is $\dfrac{{{{(4)}^n}}}{{n!}} + \dfrac{{{{( - 2)}^n}}}{{n!}}$.**

**Note:**

In the above solution we have used the law of exponent for reducing the expanding term and then we find the coefficient of ${x^n}$ in the expansion of that term which in our case is ${e^{4x}} + {e^{ - 2x}}$. Then we get the coefficients as 4 and -2. So, we get $\dfrac{{{{(4)}^n}}}{{n!}} + \dfrac{{{{( - 2)}^n}}}{{n!}}$.

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