If `C_(0),C_(1),C_(2),C_(3), . . .,C_(n)` be binomial coefficients in the expansion of `(1+x)^(n)`, then
Q. The value of the expression `C_(0)+2C_(1) +3C_(2)+. . . .+(n+1)C_(n)` is equal to

Options

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

Answer

To solve the problem, we need to evaluate the expression \( C_0 + 2C_1 + 3C_2 + \ldots + (n+1)C_n \), where \( C_r \) are the binomial coefficients from the expansion of \( (1+x)^n \).

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial coefficients \( C_r \) are given by \( C_r = \binom{n}{r} \), which represent the coefficients in the expansion of \( (1+x)^n \).

2. **Rewriting the Expression**:
The expression we want to evaluate can be rewritten as:
\[ C_0 + 2C_1 + 3C_2 + \ldots + (n+1)C_n = \sum_{r=0}^{n} (r+1) C_r \]

3. **Separating the Summation**:
We can separate the summation into two parts:
\[ \sum_{r=0}^{n} (r+1) C_r = \sum_{r=0}^{n} r C_r + \sum_{r=0}^{n} C_r \]

4. **Using the Properties of Binomial Coefficients**:
- The second summation \( \sum_{r=0}^{n} C_r \) is simply \( (1+1)^n = 2^n \).
- For the first summation \( \sum_{r=0}^{n} r C_r \), we can use the identity \( r C_r = n C_{r-1} \):
\[ \sum_{r=0}^{n} r C_r = \sum_{r=1}^{n} n C_{r-1} = n \sum_{r=0}^{n-1} C_r = n \cdot 2^{n-1} \]

5. **Combining the Results**:
Now, substituting back into our expression:
\[ \sum_{r=0}^{n} (r+1) C_r = n \cdot 2^{n-1} + 2^n \]

6. **Simplifying the Expression**:
We can factor out \( 2^{n-1} \):
\[ n \cdot 2^{n-1} + 2^n = n \cdot 2^{n-1} + 2 \cdot 2^{n-1} = (n + 2) \cdot 2^{n-1} \]

### Final Result:
Thus, the value of the expression \( C_0 + 2C_1 + 3C_2 + \ldots + (n+1)C_n \) is:
\[ (n + 2) \cdot 2^{n-1} \]