Let `P_(n)` denote the product of first n terms of the G.P. 16, 4, 1, 1/4,…., then `overset(oo)underset(n=1)Sigma (P_(n))^(1//n)`=

Answer

To solve the problem, we need to find the value of

\[ \sum_{n=1}^{\infty} (P_n)^{\frac{1}{n}} \]

where \( P_n \) is the product of the first \( n \) terms of the geometric progression (G.P.) given by 16, 4, 1, \( \frac{1}{4} \), ...

### Step 1: Identify the first term and common ratio of the G.P.

The first term \( a \) of the G.P. is 16, and the common ratio \( r \) can be calculated as:

\[ r = \frac{4}{16} = \frac{1}{4} \]

### Step 2: Write the formula for the product of the first \( n \) terms of the G.P.

The product of the first \( n \) terms of a G.P. is given by:

\[ P_n = a^n \cdot r^{\frac{n(n-1)}{2}} \]

Substituting the values of \( a \) and \( r \):

\[ P_n = 16^n \cdot \left(\frac{1}{4}\right)^{\frac{n(n-1)}{2}} \]

### Step 3: Simplify \( P_n \)

We can express \( P_n \) as follows:

\[ P_n = 16^n \cdot 4^{-\frac{n(n-1)}{2}} = 16^n \cdot (2^2)^{-\frac{n(n-1)}{2}} = 16^n \cdot 2^{-n(n-1)} = 2^{4n} \cdot 2^{-n(n-1)} = 2^{4n - n(n-1)} \]

### Step 4: Calculate \( (P_n)^{\frac{1}{n}} \)

Now we find \( (P_n)^{\frac{1}{n}} \):

\[ (P_n)^{\frac{1}{n}} = \left(2^{4n - n(n-1)}\right)^{\frac{1}{n}} = 2^{\frac{4n - n(n-1)}{n}} = 2^{4 - (n-1)} = 2^{5 - n} \]

### Step 5: Set up the series

Now we can set up the series:

\[ \sum_{n=1}^{\infty} (P_n)^{\frac{1}{n}} = \sum_{n=1}^{\infty} 2^{5-n} \]

### Step 6: Simplify the series

This series can be rewritten as:

\[ \sum_{n=1}^{\infty} 2^{5-n} = 2^5 \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n \]

### Step 7: Calculate the sum of the geometric series

The sum of the geometric series \( \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n \) is given by:

\[ \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1 \]

### Step 8: Final calculation

Thus, we have:

\[ \sum_{n=1}^{\infty} (P_n)^{\frac{1}{n}} = 2^5 \cdot 1 = 32 \]

### Final Answer

The final answer is:

\[ \sum_{n=1}^{\infty} (P_n)^{\frac{1}{n}} = 32 \]

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