If`C_0,C_1,C_2..C_n` denote the coefficients in the binomial expansion of `(1 +x)^n`, then `C_0 + 2.C_1 +3.C_2+. (n+1) C_n`

Options

A. `n2^(n-1)`
B. `(n+ 1)2^(n-1)`
C. `(n+ 2)2^(n-1)`
D. `(n+ 2) 2^(n)`

Explanation

To solve the problem, we need to find the sum \( S = C_0 + 2C_1 + 3C_2 + \ldots + (n+1)C_n \), where \( C_r \) are the coefficients in the binomial expansion of \( (1 + x)^n \).

### Step-by-step Solution:

1. **Express the Sum**:
We can express the sum \( S \) as:
\[ S = \sum_{r=0}^{n} (r + 1) C_r \]
This can be rewritten as:
\[ S = \sum_{r=0}^{n} r C_r + \sum_{r=0}^{n} C_r \]

2. **Identify the Second Sum**:
The second sum \( \sum_{r=0}^{n} C_r \) is simply the sum of the coefficients in the binomial expansion of \( (1 + x)^n \) when \( x = 1 \):
\[ \sum_{r=0}^{n} C_r = (1 + 1)^n = 2^n \]

3. **Calculate the First Sum**:
For the first sum \( \sum_{r=0}^{n} r C_r \), we can use the identity \( r C_r = n C_{r-1} \):
\[ \sum_{r=0}^{n} r C_r = \sum_{r=1}^{n} n C_{r-1} = n \sum_{r=0}^{n-1} C_r \]
The sum \( \sum_{r=0}^{n-1} C_r \) is again the sum of the coefficients in the binomial expansion of \( (1 + x)^{n-1} \) when \( x = 1 \):
\[ \sum_{r=0}^{n-1} C_r = (1 + 1)^{n-1} = 2^{n-1} \]
Therefore:
\[ \sum_{r=0}^{n} r C_r = n \cdot 2^{n-1} \]

4. **Combine the Results**:
Now, substituting back into our expression for \( S \):
\[ S = n \cdot 2^{n-1} + 2^n \]
We can factor out \( 2^{n-1} \):
\[ S = 2^{n-1} (n + 2) \]

5. **Final Result**:
Thus, the final result for the sum \( C_0 + 2C_1 + 3C_2 + \ldots + (n+1)C_n \) is:
\[ S = 2^{n-1} (n + 2) \]