Let `C_(1), C_(2), C_(3)`… are the usual binomial coefficients where `C_(r )= .^(n)C_(r )`. Let `S=C_(1)+2C_(2)+3C_(3)+…+nC_(n)`, then S is equal to

Options

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

Answer

To solve the problem, we need to find the value of the sum \( S = C_1 + 2C_2 + 3C_3 + \ldots + nC_n \), where \( C_r = \binom{n}{r} \) are the binomial coefficients.

### Step-by-Step Solution:

1. **Define the Sum**:
\[ S = C_1 + 2C_2 + 3C_3 + \ldots + nC_n \]
We can also express \( S \) starting from \( C_0 \):
\[ S = 0 \cdot C_0 + 1 \cdot C_1 + 2 \cdot C_2 + 3 \cdot C_3 + \ldots + n \cdot C_n \]

2. **Rewrite the Sum**:
We can write \( S \) as:
\[ S = 0 \cdot C_0 + 1 \cdot C_1 + 2 \cdot C_2 + \ldots + (n-1) \cdot C_{n-1} + n \cdot C_n \]

3. **Reverse the Order**:
Now, let's reverse the order of the terms:
\[ S = n \cdot C_n + (n-1) \cdot C_{n-1} + (n-2) \cdot C_{n-2} + \ldots + 1 \cdot C_1 + 0 \cdot C_0 \]

4. **Combine the Two Expressions**:
Adding the two expressions for \( S \):
\[ 2S = (0 + n)C_0 + (1 + (n-1))C_1 + (2 + (n-2))C_2 + \ldots + (n-1 + 1)C_{n-1} + (n + 0)C_n \]
This simplifies to:
\[ 2S = nC_0 + nC_1 + nC_2 + \ldots + nC_n \]

5. **Factor Out \( n \)**:
We can factor out \( n \):
\[ 2S = n(C_0 + C_1 + C_2 + \ldots + C_n) \]

6. **Use the Binomial Theorem**:
The sum of the binomial coefficients is given by:
\[ C_0 + C_1 + C_2 + \ldots + C_n = 2^n \]
Therefore, we have:
\[ 2S = n \cdot 2^n \]

7. **Solve for \( S \)**:
Dividing both sides by 2 gives:
\[ S = \frac{n \cdot 2^n}{2} = n \cdot 2^{n-1} \]

### Final Result:
Thus, the value of \( S \) is:
\[ \boxed{n \cdot 2^{n-1}} \]