Question

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

• A. 8
• B. 16
• C. 32
• D. 64

Answer 1

Answer: (C)

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