If `C_k` is the coefficient of `x^k` in the expansion of `(1 + x)^2005` and if a,d are real numbers then `sum_(k = 0)^(2005) (a + kd).C_k` =

Options

A. `(2a + 2005d)2^2004`
B. `(2a + 2005d)2^2005`
C. `(2a + 2004d)2^2005`
D. `(2a + 2004d)2^2005`

Answer

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

\[ \sum_{k=0}^{2005} (a + kd) C_k \]

where \( C_k \) is the coefficient of \( x^k \) in the expansion of \( (1 + x)^{2005} \).

### Step 1: Identify the coefficients \( C_k \)

The coefficients \( C_k \) in the expansion of \( (1 + x)^{2005} \) are given by the binomial coefficient:

\[ C_k = \binom{2005}{k} \]

### Step 2: Rewrite the summation

We can split the summation into two parts:

\[ \sum_{k=0}^{2005} (a + kd) C_k = \sum_{k=0}^{2005} a C_k + \sum_{k=0}^{2005} kd C_k \]

### Step 3: Evaluate the first summation

The first summation is:

\[ \sum_{k=0}^{2005} a C_k = a \sum_{k=0}^{2005} C_k \]

Using the binomial theorem, we know that:

\[ \sum_{k=0}^{n} C_k = (1 + 1)^{n} = 2^{2005} \]

Thus,

\[ \sum_{k=0}^{2005} a C_k = a \cdot 2^{2005} \]

### Step 4: Evaluate the second summation

For the second summation, we have:

\[ \sum_{k=0}^{2005} kd C_k = d \sum_{k=0}^{2005} k C_k \]

Using the identity \( k C_k = 2005 C_{k-1} \), we can rewrite the summation:

\[ \sum_{k=0}^{2005} k C_k = 2005 \sum_{k=1}^{2005} C_{k-1} = 2005 \sum_{j=0}^{2004} C_j \]

where we let \( j = k - 1 \). Again, using the binomial theorem:

\[ \sum_{j=0}^{2004} C_j = (1 + 1)^{2004} = 2^{2004} \]

Thus,

\[ \sum_{k=0}^{2005} k C_k = 2005 \cdot 2^{2004} \]

### Step 5: Combine the results

Now we can combine both parts:

\[ \sum_{k=0}^{2005} (a + kd) C_k = a \cdot 2^{2005} + d \cdot 2005 \cdot 2^{2004} \]

### Final Result

Factoring out \( 2^{2004} \):

\[ = 2^{2004} \left( a \cdot 2 + 2005d \right) \]

Thus, the final answer is:

\[ \sum_{k=0}^{2005} (a + kd) C_k = 2^{2004} (2a + 2005d) \]

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