Function that looks a lot like exponential, but isn't I'm looking for a continuous function f(x) with the following properties. I've been playing with exponentials, but that doesn't seem to be the answer, although my high school mathematics is a bit rusty, I must admit.


*

*$f(1) = 2$

*$f(2) = 6 = 2+4$

*$f(3) = 14 = 2+4+8$

*$f(4) = 30 = 2+4+8+16$

*And so on


I'm looking for a continuous function, so something with a meaningful answer for $f(2.5)$, which is less then $6 + \frac{14 - 6}{2} = 10$.
My simple high school math made me look at something like $f(x) = a^x$, but that doesn't seem to be the answer.
Any better ideas?
 A: If you're computer-minded, it might seem natural to express these numbers in binary, since they're sums of powers of two!
Then you see that they're:
$$f(1)=10_2$$
$$f(2)=110_2$$
$$f(3)=1110_2$$
$$f(4)=11110_2$$
and it's immediately apparent that adding $10_2$ to these will give powers of two.
A: HINT: $$\sum_{i=1}^n 2^i=2(2^n-1)$$
A: The other answers show you how to derive the function $f(x) = 2^{x+1} - 2$, but I'm just going to show how the problem is approached (in complete unrigorous terms). I know from experience how magical it seems that the above answers just managed to come up with the right function all of a sudden.
So, we look at the problem and we can see that it's forming a pattern $2, 6, 14, 30, \ldots$ - now this reminds us of something, all the numbers seem to be even, but if you look hard enough they are all $2$ less than powers of $2$. So we can see that really, the sequence is actually $$2^2 - 2, 2^3 - 2, 2^4 - 2, 2^5 - 2, \ldots$$
From this, it's pretty obvious to us that we should start thinking that this problem is related to something like $2^{\text{something}} - 2$ and once you know what to arrive at it makes getting there a whole lot easier, which is where the other answers come in.  
A: We have for the sequence $f(n)$
$$\begin{align}
f(n)&=\sum_{k=1}^n2^k\\\\
&=2^{n+1}-2\tag 1
\end{align}$$
where we summed a Geometric Progression to arrive at $(1)$.
Thus, the continuation of $f$ for real arguments is 
$$\bbox[5px,border:2px solid #C0A000]{f(x)=2^{x+1}-2}$$

NOTE:
We remark that the continuation is not unique inasmuch as we can add any continuous functions that has zeros for each integer.  As example, the function $g(x)=2^{x+1}-2+C\,\sin ( \pi x)$, where $C$ is any constant, is continuous and for all integer-valued arguments $n$, is given by $g(n)=2^{n+1}-2$ since $\sin (n\pi)=0$ for all integer values of $n$.
