Sequence And Series Question 553
Question: In the expansion of $ {\log_{e}}\frac{1}{1-x-x^{2}+x^{3}} $ , the coefficient of $ x $ is
Options:
A) 0
B) 1
C) ? 1
D) 1/2
Show Answer
Answer:
Correct Answer: B
Solution:
$ {\log_{e}}[ \frac{1}{1-x-x^{2}+x^{3}} ]={\log_{e}}[ \frac{1}{(1-x)-x^{2}(1-x)} ] $ $ ={\log_{e}}[ \frac{1}{(1-x)(1-x^{2})} ]={\log_{e}}[ \frac{1}{{{(1-x)}^{2}}(1+x)} ] $ $ ={\log_{e}}{{{(1-x)}^{-2}}{{(1+x)}^{-1}}} $ ??(i) $ ={\log_{e}}{{(1-x)}^{-2}}+{\log_{e}}{{(1+x)}^{-1}} $ $ =-2{\log_{e}}(1-x)-{\log_{e}}(1+x) $ $ =-2,[ -x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-…….\infty ] $ $ -[ x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+…..\infty ] $ , $ (\because \ -1<x\le 1) $ Hence coefficient of $ x=2-1=1 $ .
BETA
