Sequence And Series Question 572
Question: In the expansion of $ \frac{e^{7x}+e^{3x}}{e^{5x}} $ , the constant term is
Options:
A) 0
B) 1
C) 2
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{e^{7x}+e^{3x}}{e^{5x}}=e^{2x}+{e^{-2x}} $ We know $ e^{2x}=1+\frac{2x}{1\ !}+\frac{{{(2x)}^{2}}}{2\ !}+\frac{{{(2x)}^{3}}}{3\ !}+……..\infty $ ……(i) $ {e^{-2x}}=1-\frac{2x}{1\ !}+\frac{{{(2x)}^{2}}}{2\ !}-\frac{{{(2x)}^{3}}}{3\ !}+\frac{{{(2x)}^{4}}}{4\ !}-…….\infty $ ……(ii) Adding (i) and (ii), we get $ e^{2x}+{e^{-2x}}=2[ 1+\frac{{{(2x)}^{2}}}{2\ !}+\frac{{{(2x)}^{4}}}{4\ !}+……\infty ] $ Hence the constant term is 2.
BETA
