If `C_r` stands for `nC_r,` then the sum of first `(n+1)` terms of the series `a C_0-(a+d)C_1+(a+2d)C_2-(a+3d)C_3+......,` is

Answer

To solve the problem, we need to find the sum of the first \( (n+1) \) terms of the series given by:

\[ S = a C_0 - (a + d) C_1 + (a + 2d) C_2 - (a + 3d) C_3 + \ldots + (-1)^n (a + nd) C_n \]

### Step 1: Write the series in summation form
We can express the series in summation notation:

\[ S = \sum_{r=0}^{n} (-1)^r (a + rd) C_r \]

### Step 2: Separate the terms involving \( a \) and \( d \)
We can separate the series into two parts, one involving \( a \) and the other involving \( d \):

\[ S = a \sum_{r=0}^{n} (-1)^r C_r + d \sum_{r=0}^{n} (-1)^r r C_r \]

### Step 3: Evaluate the first summation
The first summation \( \sum_{r=0}^{n} (-1)^r C_r \) can be evaluated using the Binomial Theorem. It states that:

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

Thus, we have:

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

### Step 4: Evaluate the second summation
For the second summation \( \sum_{r=0}^{n} (-1)^r r C_r \), we can use the property of binomial coefficients:

\[ r C_r = n C_{r-1} \]

Thus, we can rewrite the summation as:

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

This can be simplified to:

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

Using the same reasoning as before, we find:

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

Thus, we have:

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

### Step 5: Combine results
Now substituting back into our expression for \( S \):

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

### Conclusion
The sum of the first \( (n+1) \) terms of the series is:

\[ \boxed{0} \]