Binomial Theorem And Its Simple Applications Question 17
Question: $ 1+\frac{1}{3}+\frac{1}{3}.\frac{3}{6}+\frac{1}{3}.\frac{3}{6}.\frac{5}{9}+…..\infty = $
Options:
A) $ \sqrt{\frac{2}{3}} $
B) $ \sqrt{2} $
C) $ \sqrt{3} $
D) $ \sqrt{\frac{3}{2}} $
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Answer:
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 $ ={{( 1-\frac{2}{3} )}^{-\frac{1}{2}}}={{( \frac{1}{3} )}^{-\frac{1}{2}}}=\sqrt{3} $
BETA
