MathLabs

Problem 6

Let A and E be opposite vertices of a regular octagon. A frog starts at A and jumps to an adjacent vertex until it reaches E and stops. If a_n counts paths of exactly n jumps ending at E, prove a2n−1=0a_{2n-1}=0 and a2n=(2+2)n−1−(2−2)n−12a_{2n}=\frac{(2+\sqrt{2})^{n-1}-(2-\sqrt{2})^{n-1}}{\sqrt{2}} for n=1,2,3,…n=1,2,3,\ldots.
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a2n−1=0a_{2n-1}=0
Detailed analysis

The octagon graph is bipartite and A,E have the same color, so every path from A to E has even length. Hence a2n−1=0a_{2n-1}=0.