`f(1)=1, n ge 1` `f(n+1)=2f(n)+1` then `f(n)=`

Options

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

Answer

To solve the problem, we need to find the function \( f(n) \) given the conditions \( f(1) = 1 \) and \( f(n+1) = 2f(n) + 1 \) for \( n \geq 1 \).

### Step-by-step Solution:

1. **Start with the initial condition**:
We know that \( f(1) = 1 \).

2. **Calculate \( f(2) \)**:
Using the recursive formula:
\[ f(2) = f(1 + 1) = 2f(1) + 1 = 2 \cdot 1 + 1 = 2 + 1 = 3 \]

3. **Calculate \( f(3) \)**:
Again, using the recursive formula:
\[ f(3) = f(2 + 1) = 2f(2) + 1 = 2 \cdot 3 + 1 = 6 + 1 = 7 \]

4. **Calculate \( f(4) \)**:
Continuing with the recursive formula:
\[ f(4) = f(3 + 1) = 2f(3) + 1 = 2 \cdot 7 + 1 = 14 + 1 = 15 \]

5. **Observe the pattern**:
We have calculated:
- \( f(1) = 1 \)
- \( f(2) = 3 \)
- \( f(3) = 7 \)
- \( f(4) = 15 \)

We can express these values as:
- \( f(1) = 2^1 - 1 \)
- \( f(2) = 2^2 - 1 \)
- \( f(3) = 2^3 - 1 \)
- \( f(4) = 2^4 - 1 \)

6. **Generalize the function**:
From the observed pattern, we can generalize:
\[ f(n) = 2^n - 1 \]

### Final Answer:
Thus, the function \( f(n) \) can be expressed as:
\[ f(n) = 2^n - 1 \]