A) 2
B) \[1/2\]
C) 3
D) 4
Correct Answer: B
Solution :
We have \[{{(1+x)}^{m}}=1+mx+\frac{m(m-1)}{2!}{{x}^{2}}+...\] By hypothesis, \[\frac{m(m-1)}{2}{{x}^{2}}=-\frac{1}{8}{{x}^{2}}\] Þ \[4{{m}^{2}}-4m=-1\] Þ \[{{(2m-1)}^{2}}=0\Rightarrow m=\frac{1}{2}\].You need to login to perform this action.
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