If `f(x)=(1+x)^2,` then the value of `f(x0)+f^(prime)(0)` `+(f^(0))/(2!)+(f^(0))/(3!)+(f^n(0))/(n !)dot`

Explanation

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

\[ f(0) + f'(0) + \frac{f''(0)}{2!} + \frac{f'''(0)}{3!} + \ldots + \frac{f^{(n)}(0)}{n!} \]

where \( f(x) = (1 + x)^2 \).

### Step 1: Calculate \( f(0) \)

We start by substituting \( x = 0 \) into the function:

\[ f(0) = (1 + 0)^2 = 1 \]

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

Next, we find the first derivative of \( f(x) \):

\[ f'(x) = \frac{d}{dx}[(1 + x)^2] = 2(1 + x) \cdot \frac{d}{dx}(1 + x) = 2(1 + x) \]

### Step 3: Calculate \( f'(0) \)

Now, substitute \( x = 0 \) into \( f'(x) \):

\[ f'(0) = 2(1 + 0) = 2 \]

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

Next, we find the second derivative of \( f(x) \):

\[ f''(x) = \frac{d}{dx}[f'(x)] = \frac{d}{dx}[2(1 + x)] = 2 \]

### Step 5: Calculate \( f''(0) \)

Since \( f''(x) \) is constant, we have:

\[ f''(0) = 2 \]

### Step 6: Calculate \( f'''(x) \)

Now, we find the third derivative of \( f(x) \):

\[ f'''(x) = \frac{d}{dx}[f''(x)] = \frac{d}{dx}[2] = 0 \]

### Step 7: Calculate \( f^{(n)}(x) \) for \( n \geq 3 \)

For \( n \geq 3 \):

\[ f^{(n)}(x) = 0 \]

Thus, \( f^{(n)}(0) = 0 \) for all \( n \geq 3 \).

### Step 8: Substitute into the series

Now we can substitute all these values into the series:

\[ f(0) + f'(0) + \frac{f''(0)}{2!} + \frac{f'''(0)}{3!} + \ldots + \frac{f^{(n)}(0)}{n!} \]

This becomes:

\[ 1 + 2 + \frac{2}{2!} + 0 + 0 + \ldots + 0 \]

Calculating the terms:

\[ 1 + 2 + \frac{2}{2} = 1 + 2 + 1 = 4 \]

### Final Answer

The value of the expression is:

\[ \boxed{4} \]