If `f(x)=x^(n),"n" epsilon N`, then the value of `f(1)-(f^(')(1))/(1!)+(f^(")(1))/(2!)-(f^''')(1)/(3!)+…+(-1)^(n)(f^(n)(1))/(n!)` is

Options

A. `1`
B. `2^(n)`
C. `2^(n-1)`
D. `0`

Explanation

To solve the problem, we need to evaluate the expression given in the question step by step.

### Step 1: Identify the function and its derivatives
The function is given as:
\[ f(x) = x^n \]
where \( n \) is a natural number.

Now, we compute the derivatives of \( f(x) \):
- The first derivative:
\[ f'(x) = n x^{n-1} \]
- The second derivative:
\[ f''(x) = n(n-1) x^{n-2} \]
- The third derivative:
\[ f'''(x) = n(n-1)(n-2) x^{n-3} \]
- Continuing this pattern, the \( n \)-th derivative is:
\[ f^{(n)}(x) = n(n-1)(n-2) \cdots 1 = n! \text{ (when evaluated at } x = 1) \]

### Step 2: Evaluate the derivatives at \( x = 1 \)
Now, we evaluate these derivatives at \( x = 1 \):
- \( f(1) = 1^n = 1 \)
- \( f'(1) = n \)
- \( f''(1) = n(n-1) \)
- \( f'''(1) = n(n-1)(n-2) \)
- Continuing this, we find that:
\[ f^{(k)}(1) = n(n-1)(n-2)\cdots(n-k+1) \text{ for } k \leq n \]
and \( f^{(k)}(1) = 0 \) for \( k > n \).

### Step 3: Substitute the values into the expression
We substitute these values into the expression:
\[ f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + \ldots + (-1)^n \frac{f^{(n)}(1)}{n!} \]
This becomes:
\[ 1 - \frac{n}{1!} + \frac{n(n-1)}{2!} - \frac{n(n-1)(n-2)}{3!} + \ldots + (-1)^n \frac{n!}{n!} \]

### Step 4: Recognize the pattern
The expression can be recognized as the expansion of \( (1 - 1)^n \) using the binomial theorem:
\[ (1 - x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k \]
Setting \( x = 1 \):
\[ (1 - 1)^n = 0 \]
Thus, the entire expression evaluates to \( 0 \).

### Final Answer
The value of the expression is:
\[ \boxed{0} \]