The A.M. of the series
`.^(n)C_(0), .^(n)C_(1), .^(n)C_(2),….,.^(n)C_(n)` is

Options

A. `(2^(n))/(n+1)`
B. `(2^(n))/(n)`
C. `(2^(n-1))/(n+1)`
D. `(1)/(n+1)`

Answer

To find the arithmetic mean (A.M.) of the series \( \binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n} \), we can follow these steps:

### Step 1: Identify the terms of the series
The terms of the series are the binomial coefficients:
\[ \binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n} \]

### Step 2: Calculate the sum of the terms
We know from the binomial theorem that the sum of the binomial coefficients is given by:
\[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^n \]

### Step 3: Count the number of terms
The number of terms in the series is \( n + 1 \) (from \( \binom{n}{0} \) to \( \binom{n}{n} \)).

### Step 4: Calculate the arithmetic mean
The arithmetic mean (A.M.) is calculated using the formula:
\[ \text{A.M.} = \frac{\text{Sum of terms}}{\text{Number of terms}} = \frac{2^n}{n + 1} \]

### Conclusion
Thus, the arithmetic mean of the series \( \binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n} \) is:
\[ \text{A.M.} = \frac{2^n}{n + 1} \]

### Final Answer
The A.M. of the series is \( \frac{2^n}{n + 1} \).

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