The mean of then Numbers 0,1,2,3.......n with respective weights `""^(n)C_(0),""^(n)C_(1),""^(n)C_(2)........""^(n)C_(n)` is

Options

A. `(2^(n))/(n+1)`
B. `(2^(n+1))/(n+1)`
C. `n//2`
D. None of these

Answer

To find the mean of the numbers \(0, 1, 2, \ldots, n\) with respective weights given by the binomial coefficients \(\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n}\), we will follow these steps:

### Step 1: Understand the Mean Formula
The mean of a set of numbers with weights is calculated using the formula:
\[ \text{Mean} = \frac{\sum (x_i \cdot w_i)}{\sum w_i} \]
where \(x_i\) are the numbers and \(w_i\) are their respective weights.

### Step 2: Identify the Numbers and Weights
In our case, the numbers are \(0, 1, 2, \ldots, n\) and the weights are \(\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n}\).

### Step 3: Calculate the Numerator
The numerator of the mean will be:
\[ \sum_{i=0}^{n} i \cdot \binom{n}{i} \]
This can be expanded as:
\[ 0 \cdot \binom{n}{0} + 1 \cdot \binom{n}{1} + 2 \cdot \binom{n}{2} + \ldots + n \cdot \binom{n}{n} \]

### Step 4: Use the Property of Binomial Coefficients
There is a useful identity involving binomial coefficients:
\[ \sum_{i=0}^{n} i \cdot \binom{n}{i} = n \cdot 2^{n-1} \]
This means that the sum of \(i \cdot \binom{n}{i}\) for \(i\) from \(0\) to \(n\) is equal to \(n\) times half of the total number of subsets of \(n\) elements.

### Step 5: Calculate the Denominator
The denominator is the sum of the weights:
\[ \sum_{i=0}^{n} \binom{n}{i} = 2^n \]
This is because the sum of the binomial coefficients for a given \(n\) equals \(2^n\).

### Step 6: Combine the Results
Now, substituting the results from Steps 4 and 5 into the mean formula:
\[ \text{Mean} = \frac{n \cdot 2^{n-1}}{2^n} \]
This simplifies to:
\[ \text{Mean} = \frac{n}{2} \]

### Final Answer
Thus, the mean of the numbers \(0, 1, 2, \ldots, n\) with respective weights \(\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}\) is:
\[ \frac{n}{2} \]