Suppose that `F(n +1) =( 2f(n)+1)/2` for n = 1, 2, 3,.....and f(1)= 2 Then F(101) equals = ?

Options

A. 50
B. 52
C. 54
D. none of these

Explanation

To solve the problem, we will follow these steps:

1. **Understand the recurrence relation**: We are given the function defined by the relation \( f(n + 1) = \frac{2f(n) + 1}{2} \) and the initial condition \( f(1) = 2 \).

2. **Calculate the first few terms**: We will compute the values of \( f(n) \) for the first few integers to identify a pattern.
- For \( n = 1 \):
\[ f(1) = 2 \]
- For \( n = 2 \):
\[ f(2) = \frac{2f(1) + 1}{2} = \frac{2 \cdot 2 + 1}{2} = \frac{4 + 1}{2} = \frac{5}{2} \]
- For \( n = 3 \):
\[ f(3) = \frac{2f(2) + 1}{2} = \frac{2 \cdot \frac{5}{2} + 1}{2} = \frac{5 + 1}{2} = \frac{6}{2} = 3 \]
- For \( n = 4 \):
\[ f(4) = \frac{2f(3) + 1}{2} = \frac{2 \cdot 3 + 1}{2} = \frac{6 + 1}{2} = \frac{7}{2} \]
- For \( n = 5 \):
\[ f(5) = \frac{2f(4) + 1}{2} = \frac{2 \cdot \frac{7}{2} + 1}{2} = \frac{7 + 1}{2} = \frac{8}{2} = 4 \]

3. **Identify the pattern**: From the calculations:
- \( f(1) = 2 \)
- \( f(2) = \frac{5}{2} \)
- \( f(3) = 3 \)
- \( f(4) = \frac{7}{2} \)
- \( f(5) = 4 \)

We can see that the function alternates between whole numbers and fractions, suggesting a linear pattern.

4. **Generalize the function**: We can express \( f(n) \) in terms of \( n \):
- The odd indexed terms seem to be whole numbers: \( f(1) = 2, f(3) = 3, f(5) = 4 \)
- The even indexed terms are fractions: \( f(2) = \frac{5}{2}, f(4) = \frac{7}{2} \)

We can see that:
- For odd \( n \): \( f(n) = \frac{n + 3}{2} \)
- For even \( n \): \( f(n) = \frac{n + 3}{2} \)

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

Thus, the value of \( f(101) \) is **52**.