If `C_o C_1, C_2,.......,C_n` denote the binomial coefficients in the expansion of `(1 + x)^n`, then the value of `sum_(r=0)^n (r + 1) C_r` is

Options

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

Answer

To solve the problem, we need to find the value of the sum \( S_n = \sum_{r=0}^{n} (r + 1) C_r \), where \( C_r \) are the binomial coefficients in the expansion of \( (1 + x)^n \).

### Step-by-Step Solution:

1. **Rewrite the Sum**:
We can express \( S_n \) as:
\[ S_n = \sum_{r=0}^{n} (r + 1) C_r = \sum_{r=0}^{n} r C_r + \sum_{r=0}^{n} C_r \]
This separates the sum into two parts: one involving \( r C_r \) and the other just \( C_r \).

2. **Evaluate \( \sum_{r=0}^{n} C_r \)**:
The sum \( \sum_{r=0}^{n} C_r \) is known to equal \( 2^n \) (the sum of the binomial coefficients).
\[ \sum_{r=0}^{n} C_r = 2^n \]

3. **Evaluate \( \sum_{r=0}^{n} r C_r \)**:
We can use the identity \( r C_r = n C_{r-1} \). Thus,
\[ \sum_{r=0}^{n} r C_r = \sum_{r=1}^{n} n C_{r-1} = n \sum_{r=0}^{n-1} C_r \]
The sum \( \sum_{r=0}^{n-1} C_r \) is equal to \( 2^{n-1} \) (the sum of the first \( n-1 \) binomial coefficients).
\[ \sum_{r=0}^{n} r C_r = n \cdot 2^{n-1} \]

4. **Combine the Results**:
Now we can combine the results from steps 2 and 3:
\[ S_n = n \cdot 2^{n-1} + 2^n \]

5. **Factor the Expression**:
We can factor out \( 2^{n-1} \):
\[ S_n = 2^{n-1}(n + 2) \]

### Final Answer:
Thus, the value of \( S_n \) is:
\[ S_n = 2^{n-1}(n + 2) \]