Show that one can fill a rectangle using an even number of red squares in more ways than with an odd number of such I was trying out a problem and I noticed that if a $1\times n$ rectangle is given such that it is filled using $(1\times 1$) squares of colours red, green or blue, then the number of ways it can be filled with an even number of red squares (say, $f(n)$) is $1$ more than the number of ways it can be filled using an odd number of squares (say, $g(n)).
The computations led to this observation. Firstly, we can state that $f(n)=2f(n-1)+g(n-1)$ and similarly for $g(n)$. Or otherwise, we can evaluate $f(n)$ as $\displaystyle \sum {n \choose 2j} \cdot 2^{n-2j}$ and similarly for $g(n)$. The same thing can otherwise be shown using induction as well.
I'm more curious in finding a reason to it using some combinatorial arguments instead of computation/recursion/induction or anything of this sort. 
 A: Flipping the leftmost non-green square from red to blue or blue to red changes the parity of the number of red squares. This argument would show that there are equal numbers of colourings with an even or an odd number of red squares, except that we've left out one colouring: ALL SQUARES GREEN. In this case there are no red squares, and $0$ is an even number, so even wins by one.
P.S. More generally, if $k$ colours are available for the squares, then there are $\frac{k^n+(k-2)^n}2$ ways to colour the $1\times n$ rectangle with an even number of red squares, $\frac{k^n-(k-2)^n}2$ ways with an odd number of red squares, so the difference is $(k-2)^n$ in favour of even; see this answer.
A: Let red have value $-1$, while green and blue shall have value $1$. The value of a string is the product of the values of its entries; it is $1$ when there are an even number of reds in the string, and $-1$ otherwise. If you multiply out $(-1+1+1)^n$ distributively you therefore obtain $3^n$ terms summing to $$1=(-1+1+1)^n=f(n)-g(n)\ .$$ 
