A) \[{{4}^{-2/3}}\]
B) \[\sqrt[3]{16}\]
C) \[\sqrt[3]{4}\]
D) \[{{4}^{3/2}}\]
Correct Answer: B
Solution :
Let \[S=1+\frac{2}{4}+\frac{2}{4}.\frac{5}{8}+\frac{2.5.8}{4.8.12}+...\] On comparing with \[{{(1+x)}^{n}}=1+nx+\frac{n(n-1)}{2!}{{x}^{2}}+....,\] we get \[nx=\frac{2}{4}\] ?(i) and \[\frac{n(n-1)}{2!}{{x}^{2}}=\frac{2.5}{4.8}\] From Eqs. (i) and (ii) \[\frac{\frac{n(n-1)}{2!}{{x}^{2}}}{{{n}^{2}}{{x}^{2}}}=\frac{\frac{2.5}{4.8}}{\frac{2.2}{4.4}}\] \[\Rightarrow \] \[\frac{n-1}{n}=\frac{5}{2}\] \[\Rightarrow \] \[2n-2=5n\] \[\Rightarrow \] \[n=-\frac{2}{3}\] On putting the value of n in Eq. (i), we get \[-\frac{2}{3}x=\frac{2}{4}\] \[\Rightarrow \] \[x=-\frac{3}{4}\] \[\therefore \]\[S={{(1+{{x}^{2}})}^{n}}={{\left( 1-\frac{3}{4} \right)}^{-2/3}}={{\left( \frac{1}{4} \right)}^{-2/3}}=\sqrt[3]{16}\]
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