How to solve this series $f(n) = f(n/2) + n$? Can I solve this as:
$f(n) = f(n/2) + n$
let find, $$f(n/2) = f(n/2/2) + n/2\\
  f(n/2) = f(n/4) + n/2$$
Now,
$$\begin{split}
f(n) &= f(n/4) + n/2 + n\\
f(n) &= f(n/8) + n/4 + n/2 + n
\end{split}$$
hence so on.
$$\vdots$$
Now, $n = 2^i$.
$$\begin{split}
f(2^i) &= f(2^i/2^i) + 2^i/2^{i-1} + 2^i/2^{i-2} + \cdots + 2^i\\
f(2^i) &= f(1) + 2^1 + 2^2 + \cdots + 2^i\\
f(2^i) &= 2^0 + 2^1 + 2^2 + \cdots + 2^i\\
f(2^i) &= 2^{i+1} - 1\\
f(2^i) &= 2^i\cdot 2^1 - 1\\
f(n) &= 2n - 1\\
f(2^k) &= f(2^{k-k}) + k\\
f(2^k) &= f(2^0) + k\\
 \cdots &= f(1) + k\\
 \cdots &= 1 + k
\end{split}$$
As we know
$$\begin{split}
n &= 2^k\\
\log (n) &= k \log(2)\\
k &= \log (n) / \log(2)\\
k &= \log_2 (n)\\
f(n) &= \log_2(n) + 1
\end{split}$$

At some point you need to have the value of say $f(1)$ or something. The classical way of solving this kind of questions would be to do as you did until:
$$ f(n) = f(1) + \sum_{i=0}^{k} \frac{n}{2^i}$$
where $2^k = n$. Then you should be able to solve the problem by computing $\sum_{i=0}^{k} \frac{n}{2^i}$ using geometric series.
In the case you $n$ is not a power of $2$, you must apply the reccurence formula until step $k$ with $\frac{n}{2^k} \le 1$, and the you must know what is the value of $f$ of say, interval $[0,1]$.
