A) \[\frac{2}{\sqrt{3}}\]
B) \[2\sqrt{3}\]
C) \[\frac{\sqrt{3}}{2}\]
D) \[\frac{1}{2\sqrt{3}}\]
Correct Answer: A
Solution :
| [a] As given the series is |
| \[S=1+\frac{1}{8}+\frac{1.3}{8.16}+\frac{1.3.5}{8.1.624}+.....\infty \] |
| On comparing this series with |
| \[S={{(1+x)}^{n}}=1+nx+\frac{n(n-1)}{2!}{{x}^{2}}+....\infty ,\] |
| We get |
| \[nx=\frac{1}{8}....(1)\] |
| and \[\frac{n(n-1)}{2!}{{x}^{2}}=\frac{1.3}{8.16}...(2)\] |
| From Eqs. (1) and (2), we get |
| \[\frac{\frac{n(n-1)}{2!}{{x}^{2}}}{{{n}^{2}}{{x}^{2}}}=\frac{\frac{1.3}{8.16}}{\frac{1}{8}.\frac{1}{8}}\] |
| \[\Rightarrow \frac{n-1}{2n}=\frac{3}{2}\] |
| \[\Rightarrow n-1=3n\] |
| \[\Rightarrow n=-\frac{1}{2}\] |
| On putting this value in Eq. (i) |
| \[\Rightarrow \left( -\frac{1}{2} \right)x=\frac{1}{8}\] |
| \[\Rightarrow x=-\frac{1}{4}\]. |
| But \[S={{(1+x)}^{n}}={{\left( 1-\frac{1}{4} \right)}^{-1/2}}\] |
| \[={{\left( \frac{3}{4} \right)}^{-1/2}}=\frac{2}{\sqrt{3}}.\] |
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