If n is ann integer greater than 1, then
`a-^(n)C_(1)(a-1)+.^(n)C_(2)(a-2)- . . .+(-1)^(n)(a-n)=`

Options

A. a
B. 0
C. `a^(2)`
D. `2^(n)`

Answer

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

\[ a - \binom{n}{1}(a-1) + \binom{n}{2}(a-2) - \ldots + (-1)^n \binom{n}{n}(a-n) \]

### Step-by-Step Solution:

1. **Rewrite the Expression**:
We can rewrite the expression as follows:
\[ S = a - \binom{n}{1}(a-1) + \binom{n}{2}(a-2) - \ldots + (-1)^n \binom{n}{n}(a-n) \]
This can be separated into two parts: the terms involving \(a\) and the terms involving the binomial coefficients.

2. **Factor Out \(a\)**:
Notice that each term has a common factor of \(a\):
\[ S = a \left( 1 - \binom{n}{1} + \binom{n}{2} - \ldots + (-1)^n \binom{n}{n} \right) + \text{(remaining terms)} \]

3. **Evaluate the Binomial Coefficient Sum**:
The sum \(1 - \binom{n}{1} + \binom{n}{2} - \ldots + (-1)^n \binom{n}{n}\) can be evaluated using the Binomial Theorem. Specifically, it represents the expansion of \((1 - 1)^n\):
\[ 1 - \binom{n}{1} + \binom{n}{2} - \ldots + (-1)^n \binom{n}{n} = (1 - 1)^n = 0 \]

4. **Evaluate the Remaining Terms**:
The remaining terms can be expressed as:
\[ -\left( \binom{n}{1} + 2\binom{n}{2} + \ldots + n\binom{n}{n-1} \right) \]
This can also be evaluated using the properties of binomial coefficients, but we can see that it will yield a similar alternating sum that evaluates to zero when \(x = 1\).

5. **Combine Results**:
Since both parts of our expression evaluate to zero, we have:
\[ S = a \cdot 0 + 0 = 0 \]

Thus, the final result is:
\[ \boxed{0} \]