A) \[x+3y=0\]
B) \[x-3y=0\]
C) \[3x+y=0\]
D) \[3x-y=0\]
Correct Answer: A
Solution :
\[{{\left( \frac{1-x}{1+x} \right)}^{2}}={{(1-x)}^{2}}{{(1+x)}^{-2}}\] \[=(1-2x+{{x}^{2}})[1+(-2)x+\frac{(-2)(-2-1)}{2!}{{x}^{2}}\] \[+\frac{(-2)(-2-1)(-2-2)}{3!}{{x}^{3}}+...]\] \[=(1-2x+{{x}^{2}})(1-2x+3{{x}^{2}}-4{{x}^{3}}+5{{x}^{4}}-...)\] \[\therefore \]Coefficient of\[{{x}^{4}}\]in the expansion of\[{{\left( \frac{1-x}{1+x} \right)}^{2}}\] \[=5+(-2)(-4)+(1)(3)\] \[=5+8+3=16\]
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