A) \[1-{{(1-y)}^{-1/3}}\]
B) \[1-{{(1+y)}^{1/3}}\]
C) \[1+{{(1+y)}^{-1/3}}\]
D) \[1-{{(1+y)}^{-1/3}}\]
Correct Answer: D
Solution :
We have \[y=3x+6{{x}^{2}}+10{{x}^{3}}+....\] \[\therefore \,\,1+y=1+3x+6{{x}^{2}}+10{{x}^{3}}+...\] Þ \[1+y={{(1-x)}^{-3}}\,\,\,\,\Rightarrow 1-x={{(1+y)}^{-1/3}}\] Þ \[x=1-{{(1+y)}^{-1/3}}\]
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