A) x
B) \[{{(1+x)}^{1/3}}\]
C) \[{{(1-x)}^{1/3}}\]
D) \[{{(1-x)}^{-1/3}}\]
Correct Answer: D
Solution :
[d] Let \[{{(1+y)}^{n}}=1+\frac{1}{3}x+\frac{1\times 4}{3\times 6}{{x}^{2}}+\frac{1\times 4\times 7}{3\times 6\times 9}{{x}^{3}}+...\] \[=1+ny+\frac{n(n-1)}{2!}{{y}^{2}}+...\] Comparing the terms, we get \[ny\frac{1}{3}x,\frac{n(n-1)}{2!}{{y}^{2}}=\frac{1\times 4}{3\times 6}{{x}^{2}}\] Solving \[n=-1/3,y=-x\]Hence, the given series is \[{{(1-x)}^{-1/3}}\]
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