Sequence And Series Question 232
Question: $ {2^{1/4}}{{.4}^{1/8}}{{.8}^{1/16}}{{.16}^{1/32}}………. $ is equal to
[MNR 1984; MP PET 1998; AIEEE 2002]
Options:
A) 1
B) 2
C) $ \frac{3}{2} $
D) $ \frac{5}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ {2^{1/4}}{{.4}^{1/8}}{{.8}^{1/16}}{{.16}^{1/32}}…..\infty $ $ ={2^{1/4+2/8+3/16+……}}=2^{S} $ , where $ S $ is given by $ S=\frac{1}{4}+2\frac{1}{8}+3\frac{1}{16}+4\frac{1}{32}+………..\infty $ ……(i)
$ \Rightarrow $ $ \frac{1}{2}S=\text{ }\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+……….\infty $ ……(ii) Subtracting (ii) from (i), we get $ S=1 $ . Hence required product $ =2^{1}=2 $ .
BETA
