A) x
B) \[1-x\]
C) \[1 + x\]
D) \[{{x}^{x}}\]
Correct Answer: A
Solution :
[a] We have \[y=3x+6{{x}^{2}}+10{{x}^{3}}+......\] \[\Rightarrow 1+y=1+3x+6{{x}^{2}}+10{{x}^{3}}+.....\] \[\Rightarrow 1+y={{(1-x)}^{-3}}\Rightarrow 1-x={{(1+y)}^{-1/3}}\] \[\Rightarrow x=1-{{(1+y)}^{-1/3}}\] \[=\frac{1}{3}y-\frac{1.4}{{{3}^{2}}.2}{{y}^{2}}+\frac{1.4.7}{{{3}^{2}}.3}{{y}^{3}}-........\]You need to login to perform this action.
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