If `C_(r )` stands for `""^(n)C_(r )`, then the sum of first `(n+1)` terms of the series `aC_(0)-(a+d)C_(1)+(a+2d) C_(2)-(a+3d)C_(3)+…` is

Options

A. `a//2^(n)`
B. na
C. 0
D. none of these

Answer

To find the sum of the first \( (n+1) \) terms of the series

\[ S = aC_0 - (a+d)C_1 + (a+2d)C_2 - (a+3d)C_3 + \ldots \]

we can break this series into two parts: one part involving \( a \) and the other involving \( d \).

### Step 1: Separate the terms involving \( a \) and \( d \)

We can express the series as:

\[ S = a(C_0 - C_1 + C_2 - C_3 + \ldots) + d(-C_1 + 2C_2 - 3C_3 + \ldots) \]

### Step 2: Identify the first part of the series

The first part of the series is:

\[ T_1 = C_0 - C_1 + C_2 - C_3 + \ldots + (-1)^n C_n \]

This is a well-known identity in combinatorics, which sums to 0 for \( n \geq 0 \):

\[ T_1 = 0 \]

### Step 3: Identify the second part of the series

The second part of the series is:

\[ T_2 = -C_1 + 2C_2 - 3C_3 + \ldots + (-1)^n n C_n \]

This also has a known result, which sums to 0 for \( n \geq 1 \):

\[ T_2 = 0 \]

### Step 4: Combine the results

Now we can combine the results from \( T_1 \) and \( T_2 \):

\[ S = a \cdot 0 + d \cdot 0 = 0 \]

### Final Result

Thus, the sum of the first \( (n+1) \) terms of the series is:

\[ \boxed{0} \]