Problem 136

This is a recursive function, which means the function refers back to itself to calculate subsequent terms. To solve this, we must identify the base case, $f(1)=2$. We also know that when $n$ is odd, $f(n)=f(n-2)+2$. Thus we know that $f(2017)=f(2015)+2$. Thus we know that n will always be odd in the recursion of $f(2017)$, and we add $2$ each recursive cycle, which there are $1008$ of. Thus the answer is $1008*2+2=2018$, which is answer
$\boxed{\textbf{(B)}}$.
Note that when you write out a few numbers, you find that $f(n)=n+1$ for any $n$, so $f(2017)=2018$

1. Define the function $f$ on the positive integers recursively:
$$f(1) = 2$$
$$f(n) = f(n-1) + 1 \quad \text{if } n \text{ is even}$$
$$f(n) = f(n-2) + 2 \quad \text{if } n \text{ is odd and greater than 1}$$

2. Define a new function $g(n) = f(n) - n$. We start with:
$$g(1) = f(1) - 1 = 2 - 1 = 1$$

3. Analyze the behavior of $g(n)$ for even $n$:
$$f(n) = f(n-1) + 1 \implies g(n) = f(n) - n = (f(n-1) + 1) - n = f(n-1) - (n-1) = g(n-1)$$
Therefore, $g(n) = g(n-1)$ when $n$ is even.

4. Analyze the behavior of $g(n)$ for odd $n$:
$$f(n) = f(n-2) + 2 \implies g(n) = f(n) - n = (f(n-2) + 2) - n = f(n-2) - (n-2) = g(n-2)$$
Therefore, $g(n) = g(n-2)$ when $n$ is odd.

5. From the above steps, we see that $g(n)$ is constant for all $n$. Since $g(1) = 1$, we have:
$$g(n) = 1 \quad \text{for all } n$$

6. Therefore, for any $n$:
$$f(n) = g(n) + n = 1 + n$$

7. Specifically, for $n = 2017$:
$$f(2017) = 1 + 2017 = 2018$$

The final answer is $\boxed{2018}$