If ` f(x) = x^n ` then the value of ` f(1) +(f '(1) )/(1!) +(f'' (1))/(2!) +(f' ' ' (1))/( 3!) + . . . + (f ^n (1))/( n!)` is

Options

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

Answer

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

\[ f(1) + \frac{f'(1)}{1!} + \frac{f''(1)}{2!} + \frac{f'''(1)}{3!} + \ldots + \frac{f^{(n)}(1)}{n!} \]

where \( f(x) = x^n \).

### Step 1: Calculate \( f(1) \)

We start by evaluating \( f(1) \):

\[ f(1) = 1^n = 1 \]

### Step 2: Calculate \( f'(x) \) and \( f'(1) \)

Next, we differentiate \( f(x) \):

\[ f'(x) = nx^{n-1} \]

Now, we evaluate \( f'(1) \):

\[ f'(1) = n \cdot 1^{n-1} = n \]

### Step 3: Calculate \( f''(x) \) and \( f''(1) \)

Now, we differentiate \( f'(x) \):

\[ f''(x) = n(n-1)x^{n-2} \]

Evaluating \( f''(1) \):

\[ f''(1) = n(n-1) \cdot 1^{n-2} = n(n-1) \]

### Step 4: Calculate \( f'''(x) \) and \( f'''(1) \)

Next, we differentiate \( f''(x) \):

\[ f'''(x) = n(n-1)(n-2)x^{n-3} \]

Evaluating \( f'''(1) \):

\[ f'''(1) = n(n-1)(n-2) \cdot 1^{n-3} = n(n-1)(n-2) \]

### Step 5: Generalize \( f^{(k)}(x) \)

Continuing this process, we can see a pattern emerging. The \( k \)-th derivative of \( f(x) \) is:

\[ f^{(k)}(x) = n(n-1)(n-2)\cdots(n-k+1)x^{n-k} \]

Evaluating at \( x = 1 \):

\[ f^{(k)}(1) = n(n-1)(n-2)\cdots(n-k+1) \]

### Step 6: Write the expression

Now we can write the entire expression:

\[ 1 + \frac{n}{1!} + \frac{n(n-1)}{2!} + \frac{n(n-1)(n-2)}{3!} + \ldots + \frac{n(n-1)(n-2)\cdots(n-n)}{n!} \]

The last term \( \frac{n(n-1)(n-2)\cdots(n-n)}{n!} \) simplifies to \( \frac{n!}{n!} = 1 \).

### Step 7: Recognize the series as a binomial expansion

The expression can be recognized as the expansion of \( (1 + 1)^n \):

\[ 1 + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^n \]

### Conclusion

Thus, the value of the original expression is:

\[ \boxed{2^n} \]