A) \[\sqrt{\frac{2}{3}}\]
B) \[\sqrt{2}\]
C) \[\sqrt{3}\]
D) \[\sqrt{\frac{3}{2}}\]
Correct Answer: C
Solution :
[c] Let the given series be the expansion of \[{{(1+x)}^{n}}\], then it is identical with \[1+nx+\frac{n(n-1)}{2!}.{{x}^{2}}+.....\] \[\therefore \,\,nx=\frac{1}{3}...(1)\] \[\frac{n(n-1)}{2}.{{x}^{2}}=\frac{1}{6}...(2)\] Solving the equations (1) and (2) we get \[n=-\frac{1}{2}\] and \[x=-\frac{2}{3}\] \[\therefore \] The given series \[={{\left( 1-\frac{2}{3} \right)}^{-\frac{1}{2}}}={{\left( \frac{1}{3} \right)}^{-\frac{1}{2}}}=\sqrt{3}\]
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