A) 0
B) 1/3
C) 2/3
D) 1/6
Correct Answer: B
Solution :
\[({{e}^{x}}-1)({{e}^{-x}}+1)=({{e}^{x}}-1)\left( \frac{1+{{e}^{x}}}{{{e}^{x}}} \right)\] \[=\frac{{{e}^{2x}}-1}{{{e}^{x}}}={{e}^{x}}-{{e}^{-x}}=2\left\{ x+\frac{{{x}^{3}}}{3!}+.... \right\}\] \ The coefficient of\[{{x}^{3}}=2.\frac{1}{3!}=\frac{1}{3}\].You need to login to perform this action.
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