A) x
B) \[{{(1+x)}^{1/3}}\]
C) \[{{(1-x)}^{1/3}}\]
D) \[{{(1-x)}^{-1/3}}\]
Correct Answer: D
Solution :
Let \[{{(1+y)}^{n}}=1+\frac{1}{3}x+\frac{1.4}{3.6}{{x}^{2}}+\frac{1.4.7}{3.6.9}{{x}^{3}}+....\] \[=1+ny+\frac{n(n-1)}{2!}{{y}^{2}}+.....\] Comparing the terms, we get \[ny=\frac{1}{3}x,\frac{n(n-1)}{2!}{{y}^{2}}=\frac{1.4}{3.6}{{x}^{2}}\] Solving, \[n=-\frac{1}{3},y=-x\]. Hence given series \[={{(1-x)}^{-1/3}}\]
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