Sequence And Series Question 370
Question: If $ S_{n}= $ $ (1+{3^{-1}})(1+{3^{-2}})(1+{3^{-4}})(1+{3^{-8}})….(1+{3^{-2^{n}}}), $ then $ {S_{\infty }} $ is equal to
Options:
A) 1
B) $ \frac{1}{2} $
C) $ \frac{3}{2} $
D) None
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ S_{n}=(1+{3^{-1}})(1+{3^{-2}})(1+{3^{-4}})(1+{3^{-4}})….(1+{3^{-2n}}) $
$ \Rightarrow (1-{3^{-1}})S_{n} $ $ =(1-{3^{-1}})(1+{3^{-1}})(1+{3^{-2}})(1+{3^{-4}})(1+{3^{-8}})….(1+{3^{-2^{n}}}) $
$ \Rightarrow \frac{2}{3}S_{n}=(1-{3^{-2}})(1+{3^{-2}})(1+{3^{-4}})(1+{3^{-8}})….(1+{3^{-2n}}) $ $ =(1-{3^{-4}})(1+{3^{-4}})(1+{3^{-8}})….(1+{3^{-2^{n}}}) $ $ =(1-{3^{-8}})(1+{3^{-8}})….(1+{3^{-2^{n}}}) $ $ =(1-{3^{-2^{n}}})(1+{3^{-2^{n}}})=1-{{({3^{-2^{n}}})}^{2}}=1-{3^{-{2^{n+1}}}} $
$ \Rightarrow S_{n}=\frac{3}{2}(1-{3^{-{2^{n+1}}}}) $
$ \therefore {S_{\infty }}=\frac{3}{2}(1-0)=\frac{3}{2} $
BETA
