Since an, bn, an+1 form an arithmetic sequence, we can write the equation $$2b_{n}=2^{n}+2^{n+1}$$.

Solving for bn, we get $$b_{n}= \frac {3}{2}\cdot 2^{n}$$.

Now, to find the sum of the first n terms, Sn, of the sequence {bn}, we have:

$$S_{n}= \frac {3}{2}(2^{1}+2^{2}+2^{3}+…+2^{n})$$.

Recognizing that this is a geometric series, we can use the formula for the sum of a geometric series:

$$S_{n}=\frac{a_1(1-r^n)}{1-r}$$

where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. In our case, $a_1=\frac{3}{2}\cdot2^1=3$, $r=2$, and $n$ is the number of terms.

Substituting these values into the formula, we get:

$$S_{n}=\frac{3(1-2^n)}{1-2}=3(2^n-1)$$

Hence, the sum of the first n terms of the sequence {bn} is \boxed{S_n=3(2^n-1)}.