Question
The co-efficient of $$x^n$$ in the expansion of $$\frac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}}$$ is
A.
$$\frac{{{4^{n - 1}} + {{\left( { - 2} \right)}^n}}}{{n!}}$$
B.
$$\frac{{{4^{n - 1}} + {2^n}}}{{n!}}$$
C.
$$\frac{{{4^{n}} + {{\left( { - 2} \right)}^n}}}{{n!}}$$
D.
$$\frac{{{4^{n - 1}} + {{\left( { - 2} \right)}^{n - 1}}}}{{n!}}$$
Answer :
$$\frac{{{4^{n}} + {{\left( { - 2} \right)}^n}}}{{n!}}$$
Solution :
$$\eqalign{
& \frac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}} = {e^{4x}} + {e^{ - 2x}} \cr
& = \left[ {1 + 4x + \frac{{{{\left( {4x} \right)}^2}}}{{2!}} + .....} \right] + \left[ {1 + \left( { - 2x} \right) + \frac{{{{\left( { - 2x} \right)}^2}}}{{2!}} + .....} \right] \cr
& \therefore {\text{coeff}}{\text{. of }}{x^n} = \frac{{{4^n}}}{{n!}} + \frac{{{{\left( { - 2} \right)}^n}}}{{n!}} \cr} $$