If `.^nC_0,^nC_1,^nC_2,.....^nC_n` are frequencies of `n+1` observations `1,2,2^2,.....2^n` such that mean is `729/2^n` then value of n is:

Answer

To solve the problem, we need to find the value of \( n \) given that the frequencies of the observations \( 1, 2, 2^2, \ldots, 2^n \) are \( \binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n} \) respectively, and that the mean of these observations is \( \frac{729}{2^n} \).

### Step 1: Set up the observations and frequencies
We have the observations:
- \( x_0 = 1 \)
- \( x_1 = 2 \)
- \( x_2 = 2^2 \)
- ...
- \( x_n = 2^n \)

The corresponding frequencies are:
- \( f_0 = \binom{n}{0} \)
- \( f_1 = \binom{n}{1} \)
- \( f_2 = \binom{n}{2} \)
- ...
- \( f_n = \binom{n}{n} \)

### Step 2: Calculate the mean
The mean \( \bar{x} \) is given by the formula:
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]
Where \( \sum f_i = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^n \) (by the binomial theorem).

Now, we calculate \( \sum f_i x_i \):
\[ \sum f_i x_i = 1 \cdot \binom{n}{0} + 2 \cdot \binom{n}{1} + 2^2 \cdot \binom{n}{2} + \ldots + 2^n \cdot \binom{n}{n} \]
This can be simplified using the binomial theorem:
\[ \sum f_i x_i = (1 + 2)^n = 3^n \]

### Step 3: Substitute into the mean formula
Now substituting back into the mean formula:
\[ \bar{x} = \frac{3^n}{2^n} \]

### Step 4: Set the mean equal to the given value
We are given that the mean is \( \frac{729}{2^n} \):
\[ \frac{3^n}{2^n} = \frac{729}{2^n} \]

### Step 5: Cancel \( 2^n \) and solve for \( n \)
Cancelling \( 2^n \) from both sides gives:
\[ 3^n = 729 \]

### Step 6: Express 729 as a power of 3
We know that \( 729 = 3^6 \), so:
\[ 3^n = 3^6 \]

### Step 7: Equate the exponents
This implies:
\[ n = 6 \]

### Conclusion
Thus, the value of \( n \) is \( \boxed{6} \).