Question

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

Answer 1

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