Sequence And Series Question 488
Question: In the expansion of $ \frac{1-2x+3x^{2}}{e^{x}} $ , the coefficient of $ x^{5} $ will be
Options:
A) $ \frac{71}{120} $
B) $ -\frac{71}{120} $
C) $ \frac{31}{40} $
D) $ -\frac{31}{40} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ (1-2x+3x^{2}){e^{-x}} $ $ =(1-2x+3x^{2}){ 1-\frac{x}{1\ !}+\frac{x^{2}}{2\ !}-\frac{x^{3}}{3\ !}+……. } $
$ \therefore $ The coefficient of $ x^{5} $ $ =1( -\frac{1}{5\ !} )+(-2),( \frac{1}{4\ !} )+3( -\frac{1}{3\ !} ) $ $ =1+( \frac{1}{1.2}-\frac{1}{2.3} )+( \frac{1}{3.4}-\frac{1}{4.5} )+…. $ .
BETA
