Number of faces in a planar graph bounded by odd length cycles?

Question

Suppose that every face in a planar graph is bounded by odd length cycles, then the number of faces of this planar graph is even. I want to prove this using Euler's formula, but not really sure where to start.

Answer 1

Use the following: $$2E = \sum_{k=1}^\infty (2k+1) F_{2k+1}$$

Since every odd length cycle face contributes its edges twice; by Euler's formula $V+F-E=2$ we have:

$$2V+2F-4 = 2E$$ $$2\left(V+F-2\right)=\left(\sum_{k=1}^\infty (2k+1)F_{2k+1}\right) $$

The left hand side is even. The right hand side is odd, when an odd number of odd length cycle faces appear, so it has to be even...