Domino's!

For the $2\times n$ case there are $F_{n+1}$ ways to cover the rectangle where $F_{n}$ denotes the $n^{th}$ Fibonacci number. So for the given case the number of tilings is $F_{16}=\boxed{987}$

We will prove the general case by setting up a recurrence relation and showing that it is the same as the recurrence relation for the Fibonacci numbers. Then we just need a little care thinking about the initial values and we will have a proof.

Let $T_{n}$ be the number of ways to tile the $2\times n$ rectangle. Imagine the $2\times n$ tiling set out in a long horizontal rectangle, and being built up from left to right. There are only two ways it can finish - with a vertical domino or with two horizontal dominos stacked above each other. This observation and a little thought gives us

$T_{n}=T_{n-1}+T_{n-2}$

which is the recurrence relation for the Fibonacci numbers!

It is easy to see that $T_{1}=1$ and $T_{2}=2$

and so our tiling sequence begins 1,2,3,5,8....

The Fibonacci sequence begins 1,1,2,3,5,8...

So we can see that $\boxed{T_{n}=F_{n+1}}$