If `f(n+1)=(2f(n)+1)/(2)` for `n=1, 2, 3………..` and `f(1)=2`, then `(f(101))/(10)` is equal to

Answer

To solve the problem, we need to find the value of \( \frac{f(101)}{10} \) given the recurrence relation \( f(n+1) = \frac{2f(n) + 1}{2} \) and the initial condition \( f(1) = 2 \).

### Step-by-Step Solution:

1. **Identify the recurrence relation**:
The recurrence relation is given as:
\[ f(n+1) = \frac{2f(n) + 1}{2} \]
with the initial condition \( f(1) = 2 \).

2. **Calculate subsequent values of \( f(n) \)**:
We will calculate the values of \( f(n) \) for \( n = 1, 2, 3, \ldots, 101 \).

- For \( n = 1 \):
\[ f(2) = \frac{2f(1) + 1}{2} = \frac{2 \cdot 2 + 1}{2} = \frac{4 + 1}{2} = \frac{5}{2} \]

- For \( n = 2 \):
\[ f(3) = \frac{2f(2) + 1}{2} = \frac{2 \cdot \frac{5}{2} + 1}{2} = \frac{5 + 1}{2} = \frac{6}{2} = 3 \]

- For \( n = 3 \):
\[ f(4) = \frac{2f(3) + 1}{2} = \frac{2 \cdot 3 + 1}{2} = \frac{6 + 1}{2} = \frac{7}{2} \]

- For \( n = 4 \):
\[ f(5) = \frac{2f(4) + 1}{2} = \frac{2 \cdot \frac{7}{2} + 1}{2} = \frac{7 + 1}{2} = \frac{8}{2} = 4 \]

- For \( n = 5 \):
\[ f(6) = \frac{2f(5) + 1}{2} = \frac{2 \cdot 4 + 1}{2} = \frac{8 + 1}{2} = \frac{9}{2} \]

Continuing this process, we can see a pattern forming.

3. **Recognize the pattern**:
The values of \( f(n) \) appear to be alternating between integers and half-integers. Specifically, we can observe:
- \( f(1) = 2 \)
- \( f(2) = \frac{5}{2} \)
- \( f(3) = 3 \)
- \( f(4) = \frac{7}{2} \)
- \( f(5) = 4 \)
- \( f(6) = \frac{9}{2} \)
- \( f(7) = 5 \)
- \( f(8) = \frac{11}{2} \)
- \( f(9) = 6 \)
- \( f(10) = \frac{13}{2} \)
- Continuing this, we can see that \( f(n) \) can be expressed as:
- For odd \( n \): \( f(n) = \frac{n + 3}{2} \)
- For even \( n \): \( f(n) = \frac{n + 2}{2} \)

4. **Calculate \( f(101) \)**:
Since \( 101 \) is odd:
\[ f(101) = \frac{101 + 3}{2} = \frac{104}{2} = 52 \]

5. **Calculate \( \frac{f(101)}{10} \)**:
\[ \frac{f(101)}{10} = \frac{52}{10} = 5.2 \]

### Final Answer:
\[ \frac{f(101)}{10} = 5.2 \]