Question

Let $a _{ n }=16,4,1, \ldots$ be a geometric sequence. Define $P _{ n }$ as the product of the first $n$ terms. The value of $\displaystyle\sum_{n=1}^{\infty} \sqrt[n]{P_n}$

• A. 8
• B. 16
• C. 32
• D. 64

Answer 1

Answer: (C)

For the G.P. a, ar, $a^2{ }^2, \ldots \ldots . . . .$.
$P _{ n }= a (a r)\left(a r^2\right) \ldots \ldots . .\left(a^{ n -1}\right)= a ^{ n } r ^{ n ( n -1) / 2}$
$\therefore S =\displaystyle\sum_{ n =1}^{\infty} \sqrt[ p ]{ P _{ n }}=\displaystyle\sum_{ n =1}^{\infty} ar ^{( n -1) / 2}$
now, $ \displaystyle\sum_{ n =1}^{\infty} ar ^{( n -1) / 2}= a [1+\sqrt{ r }+ r + r \sqrt{ r }+\ldots \ldots+\infty]=\frac{ a }{1-\sqrt{ r }}$
Given $a =16$ and $r =1 / 4$
$\therefore S =\frac{16}{1-(1 / 2)}=32$