Let `a_(n)=16,4,1,………..` be a geometric sequence. The value of `Sigma_(n=1)^(oo)rootn(P_(n))`, where `P_(n)` is the product of the first n terms, is equal to.

Options

A. 8
B. 16
C. 32
D. 64

Answer

To solve the problem, we need to analyze the geometric sequence given and calculate the required summation. Let's break it down step by step.

### Step 1: Identify the geometric sequence
The sequence given is \( a_n = 16, 4, 1, \ldots \). This is a geometric sequence where:
- The first term \( a = 16 \)
- The common ratio \( r = \frac{4}{16} = \frac{1}{4} \)

### Step 2: Write the formula for the product of the first n terms
The product of the first \( n \) terms of a geometric sequence can be expressed as:
\[ P_n = a^n \cdot r^{\frac{n(n-1)}{2}} \]
For our sequence:
\[ P_n = 16^n \cdot \left(\frac{1}{4}\right)^{\frac{n(n-1)}{2}} \]

### Step 3: Simplify the expression for \( P_n \)
We can rewrite \( P_n \):
\[ P_n = 16^n \cdot 4^{-\frac{n(n-1)}{2}} = 16^n \cdot (2^2)^{-\frac{n(n-1)}{2}} = 16^n \cdot 2^{-n(n-1)} = 2^{4n} \cdot 2^{-n(n-1)} = 2^{4n - n(n-1)} \]
Thus,
\[ P_n = 2^{4n - n(n-1)} = 2^{4n - n^2 + n} = 2^{5n - n^2} \]

### Step 4: Find the nth root of \( P_n \)
Now, we need to find \( \sqrt[n]{P_n} \):
\[ \sqrt[n]{P_n} = \sqrt[n]{2^{5n - n^2}} = 2^{\frac{5n - n^2}{n}} = 2^{5 - n} \]

### Step 5: Set up the summation
We need to calculate:
\[ \sum_{n=1}^{\infty} \sqrt[n]{P_n} = \sum_{n=1}^{\infty} 2^{5 - n} \]
This can be rewritten as:
\[ \sum_{n=1}^{\infty} 2^{5} \cdot 2^{-n} = 32 \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n} \]

### Step 6: Calculate the infinite geometric series
The sum of an infinite geometric series \( \sum_{n=0}^{\infty} ar^n \) is given by \( \frac{a}{1 - r} \), where \( |r| < 1 \). Here, \( a = \frac{1}{2} \) and \( r = \frac{1}{2} \):
\[ \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1 \]

### Step 7: Final calculation
Thus, we have:
\[ \sum_{n=1}^{\infty} \sqrt[n]{P_n} = 32 \cdot 1 = 32 \]

### Final Answer
The value of \( \Sigma_{n=1}^{\infty} \sqrt[n]{P_n} \) is \( \boxed{32} \).