The arithmetic mean of `""^(n)C_(0),""^(n)C_(1),""^(n)C_(2), ..., ""^(n)C_(n)`, is

Options

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

Answer

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

### Step 1: Identify the total number of terms
The binomial coefficients \( \binom{n}{k} \) for \( k = 0, 1, 2, \ldots, n \) gives us a total of \( n + 1 \) terms.

### Step 2: Calculate the sum of the binomial coefficients
We know from the binomial theorem that:
\[ (1 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} \]
This simplifies to:
\[ 2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} \]
Thus, the sum of the binomial coefficients is \( 2^n \).

### Step 3: Calculate the arithmetic mean
The arithmetic mean \( A \) is given by the formula:
\[ A = \frac{\text{Sum of terms}}{\text{Number of terms}} \]
Substituting the values we found:
\[ A = \frac{2^n}{n + 1} \]

### Final Answer
The arithmetic mean of \( \binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n} \) is:
\[ \frac{2^n}{n + 1} \]

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