A) \[\sqrt{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[\sqrt{3}\]
D) \[\frac{1}{\sqrt{3}}\]
Correct Answer: A
Solution :
Let the given series be identical with the expansion of \[{{(1+x)}^{n}}\] i.e. with \[1+nx+\frac{n(n-1)}{2\,!}{{x}^{2}}+....;\,|x|\,<\,1\]. Then, \[nx=\frac{1}{4}\] and \[\frac{n(n-1)}{2}{{x}^{2}}=\frac{1}{4}\,.\,\frac{3}{8}=\frac{3}{32}\] Solving these two equations for n and x. We get \[x=-\frac{1}{2}\] and \[n=-\frac{1}{2}\]. \ Sum of the given series = \[{{(1+x)}^{n}}={{\left( 1-\frac{1}{2} \right)}^{-1/2}}={{2}^{1/2}}=\sqrt{2.}\]
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