If `C_(0),C_(1),C_(2),C_(3), . . .,C_(n)` be binomial coefficients in the expansion of `(1+x)^(n)`, then
Q. The value of the expression `C_(0)-2C_(1)+3C_(2)-. . . .+(-1)^(n)(n+1)C_(n)` is equal to

Options

A. 0
B. `2^(n)(n+3)`
C. `2^(n-1)(n-2)`
D. None of these

Answer

To solve the expression \( C_0 - 2C_1 + 3C_2 - \ldots + (-1)^n (n+1)C_n \), where \( C_r = \binom{n}{r} \) are the binomial coefficients from the expansion of \( (1+x)^n \), we can follow these steps:

### Step 1: Rewrite the expression
We can express the given sum in a more manageable form:
\[ S_n = \sum_{r=0}^{n} (-1)^r (r+1) C_r \]
This can be separated into two sums:
\[ S_n = \sum_{r=0}^{n} (-1)^r C_r + \sum_{r=0}^{n} (-1)^r r C_r \]

### Step 2: Evaluate the first sum
The first sum \( \sum_{r=0}^{n} (-1)^r C_r \) can be evaluated using the Binomial Theorem:
\[ \sum_{r=0}^{n} (-1)^r \binom{n}{r} = (1-1)^n = 0 \]

### Step 3: Evaluate the second sum
For the second sum \( \sum_{r=0}^{n} (-1)^r r C_r \), we can use the identity \( r C_r = n C_{r-1} \):
\[ \sum_{r=0}^{n} (-1)^r r C_r = n \sum_{r=1}^{n} (-1)^r C_{r-1} \]
Changing the index of summation:
\[ = n \sum_{s=0}^{n-1} (-1)^{s+1} C_s = -n \sum_{s=0}^{n-1} (-1)^s C_s \]
Using the result from Step 2:
\[ \sum_{s=0}^{n-1} (-1)^s C_s = (1-1)^{n-1} = 0 \]

### Step 4: Combine the results
Now we combine the results from both sums:
\[ S_n = 0 - n \cdot 0 = 0 \]

### Final Answer
Thus, the value of the expression \( C_0 - 2C_1 + 3C_2 - \ldots + (-1)^n (n+1)C_n \) is:
\[ \boxed{0} \]