If `n in N, n > 1`, then value of `E= a - ""^(n)C_(1) (a-1) + ""^(n)C_(2) (a -2)+ ... + (- 1)^(n) (a-n) (""^(n)C_(n))` is

Options

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

Answer

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

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

### Step 1: Rewrite the Expression
We can rewrite the expression \(E\) using summation notation:

\[ E = \sum_{r=0}^{n} (-1)^r (a - r) \binom{n}{r} \]

### Step 2: Split the Summation
We can split the summation into two parts:

\[ E = \sum_{r=0}^{n} (-1)^r a \binom{n}{r} - \sum_{r=0}^{n} (-1)^r r \binom{n}{r} \]

### Step 3: Evaluate the First Summation
The first summation can be simplified:

\[ \sum_{r=0}^{n} (-1)^r a \binom{n}{r} = a \sum_{r=0}^{n} (-1)^r \binom{n}{r} = a(1 - 1)^n = a \cdot 0 = 0 \]

### Step 4: Evaluate the Second Summation
For the second summation, we can use the identity \(r \binom{n}{r} = n \binom{n-1}{r-1}\):

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

This can be simplified further:

\[ = -n \sum_{s=0}^{n-1} (-1)^s \binom{n-1}{s} = -n(1 - 1)^{n-1} = -n \cdot 0 = 0 \]

### Step 5: Combine the Results
Now, substituting back into our expression for \(E\):

\[ E = 0 - 0 = 0 \]

### Conclusion
Thus, the value of \(E\) is:

\[ \boxed{0} \]