if `f(x) = x^n` then the value of `f(1) - (f'(1))/(1!) + (f''(1))/(2!) + ---+((-1)^n f''^--n times (1))/(n!)`

Options

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

Answer

To solve the problem, we need to evaluate the expression given \( f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + \ldots + \frac{(-1)^n f^{(n)}(1)}{n!} \) where \( f(x) = x^n \).

### Step-by-step Solution:

1. **Calculate \( f(1) \)**:
\[ f(1) = 1^n = 1 \]

2. **Calculate \( f'(x) \)**:
\[ f'(x) = \frac{d}{dx}(x^n) = n x^{n-1} \]
Now, substituting \( x = 1 \):
\[ f'(1) = n \cdot 1^{n-1} = n \]

3. **Calculate \( f''(x) \)**:
\[ f''(x) = \frac{d^2}{dx^2}(x^n) = n(n-1)x^{n-2} \]
Now, substituting \( x = 1 \):
\[ f''(1) = n(n-1) \cdot 1^{n-2} = n(n-1) \]

4. **Calculate \( f'''(x) \)**:
\[ f'''(x) = \frac{d^3}{dx^3}(x^n) = n(n-1)(n-2)x^{n-3} \]
Now, substituting \( x = 1 \):
\[ f'''(1) = n(n-1)(n-2) \cdot 1^{n-3} = n(n-1)(n-2) \]

5. **Continue this process up to \( f^{(n)}(x) \)**:
The \( n \)-th derivative is:
\[ f^{(n)}(x) = n(n-1)(n-2)(n-3)\ldots(1) = n! \]
Now, substituting \( x = 1 \):
\[ f^{(n)}(1) = n! \]

6. **Substituting all values back into the expression**:
We substitute \( f(1) \), \( f'(1) \), \( f''(1) \), ..., \( f^{(n)}(1) \) into the original expression:
\[ y = 1 - \frac{n}{1!} + \frac{n(n-1)}{2!} - \frac{n(n-1)(n-2)}{3!} + \ldots + \frac{(-1)^n n!}{n!} \]

7. **Recognizing the pattern**:
This expression resembles the expansion of \( (1 + x)^n \) evaluated at \( x = -1 \):
\[ (1 - 1)^n = 0^n = 0 \]
This means that:
\[ y = 1 - 1^n = 1 - 1 = 0 \]

### Final Result:
Thus, the value of the expression is:
\[ \boxed{0} \]