If `C_(0), C_(1), C_(2), ..., C_(n)` denote the binomial
cefficients in the expansion of `(1 + x )^(n)` , then
` a C_(0) + (a + b) C_(1) + (a + 2b) C_(2) + ...+ (a + nb)C_(n) = `.

Options

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

Answer

To solve the problem, we need to evaluate the expression:

\[ S_n = a C_0 + (a + b) C_1 + (a + 2b) C_2 + \ldots + (a + nb) C_n \]

where \( C_r \) are the binomial coefficients from the expansion of \( (1 + x)^n \).

### Step 1: Rewrite the expression
We can break down the expression \( S_n \) into two separate sums:

\[ S_n = \sum_{r=0}^{n} a C_r + \sum_{r=0}^{n} (r b) C_r \]

### Step 2: Evaluate the first sum
The first sum can be simplified as follows:

\[ \sum_{r=0}^{n} a C_r = a \sum_{r=0}^{n} C_r \]

Using the binomial theorem, we know that:

\[ \sum_{r=0}^{n} C_r = 2^n \]

Thus,

\[ \sum_{r=0}^{n} a C_r = a \cdot 2^n \]

### Step 3: Evaluate the second sum
For the second sum, we can use the identity \( r C_r = n C_{r-1} \):

\[ \sum_{r=0}^{n} (r b) C_r = b \sum_{r=0}^{n} r C_r = b \sum_{r=1}^{n} n C_{r-1} \]

Changing the index of summation:

\[ \sum_{r=1}^{n} n C_{r-1} = n \sum_{s=0}^{n-1} C_s = n \cdot 2^{n-1} \]

Thus, we have:

\[ \sum_{r=0}^{n} (r b) C_r = b \cdot n \cdot 2^{n-1} \]

### Step 4: Combine the results
Now we can combine both sums:

\[ S_n = a \cdot 2^n + b \cdot n \cdot 2^{n-1} \]

### Step 5: Factor out common terms
Factoring out \( 2^{n-1} \):

\[ S_n = 2^{n-1} (2a + bn) \]

### Final Result
Thus, the final answer is:

\[ S_n = 2^{n-1} (2a + bn) \]