If `C_(1), C_(2), C_(3),…C_(n)` denote the coefficients in the binomial expansion of `(1+x)^(n)`, then the value of `""^(n)C_(1) + 2.""^(n)C_(2) + 3.""^(n)C_(3) + ….+ n. ""^(n)C_(n)` is

Options

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

Answer

To solve the problem, we need to find the value of the expression:

\[ 1 \cdot C_1 + 2 \cdot C_2 + 3 \cdot C_3 + \ldots + n \cdot C_n \]

where \( C_k = \binom{n}{k} \) are the coefficients in the binomial expansion of \( (1+x)^n \).

### Step-by-Step Solution:

1. **Write the Binomial Expansion**:
The binomial expansion of \( (1+x)^n \) is given by:
\[ (1+x)^n = \sum_{k=0}^{n} C_k x^k = \sum_{k=0}^{n} \binom{n}{k} x^k \]

2. **Differentiate the Expansion**:
To find the weighted sum \( 1 \cdot C_1 + 2 \cdot C_2 + \ldots + n \cdot C_n \), we differentiate the binomial expansion with respect to \( x \):
\[ \frac{d}{dx}[(1+x)^n] = n(1+x)^{n-1} \]
On the right side, differentiating the series gives:
\[ \sum_{k=1}^{n} k \cdot C_k x^{k-1} = \sum_{k=1}^{n} k \cdot \binom{n}{k} x^{k-1} \]

3. **Substitute \( x = 1 \)**:
Now, we substitute \( x = 1 \) into both sides:
\[ n(1+1)^{n-1} = n \cdot 2^{n-1} \]
For the series, substituting \( x = 1 \) gives:
\[ \sum_{k=1}^{n} k \cdot C_k = 1 \cdot C_1 + 2 \cdot C_2 + \ldots + n \cdot C_n \]

4. **Combine Results**:
Therefore, we have:
\[ 1 \cdot C_1 + 2 \cdot C_2 + \ldots + n \cdot C_n = n \cdot 2^{n-1} \]

### Final Result:
Thus, the value of \( 1 \cdot C_1 + 2 \cdot C_2 + 3 \cdot C_3 + \ldots + n \cdot C_n \) is:
\[ \boxed{n \cdot 2^{n-1}} \]