Sequence And Series Question 372
Question: What is the sum of the series $ 1+\frac{1}{8}+\frac{1.3}{8.16}+\frac{1.3.5}{8.16.24}+….\infty $ ?
Options:
A) $ \frac{2}{\sqrt{3}} $
B) $ 2\sqrt{3} $
C) $ \frac{\sqrt{3}}{2} $
D) $ \frac{1}{2\sqrt{3}} $
Show Answer
Answer:
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 ( -\frac{1}{2} )x=\frac{1}{8} $
$ \Rightarrow x=-\frac{1}{4} $ . But $ S={{(1+x)}^{n}}={{( 1-\frac{1}{4} )}^{-1/2}} $ $ ={{( \frac{3}{4} )}^{-1/2}}=\frac{2}{\sqrt{3}}. $
BETA
