ABCDEFGH is a regular octagon. A and E are opposite vertices of the octagon. A frog starts jumping from vertex to vertex, beginning from A. From any vertex of the octagon except E, it may jump to either of the two adjacent vertices. When it reaches E, the frog stops and stays there. Let an be the number of distinct paths of exactly n jumps ending in E. Then what is the value of a2n – 1?
Explanation:
The frog has to jump at least four times to reach E.
i.e. a4 = 2
(i.e. (i) A-B, B-C,C-D and D-E; (ii) A-H, H-G, G-F and F-E).
If the frog keeps jumping on left (right) hand side vertices it will not take more than four jumps.
If it jumps on right vertex and then keeps on jumping left vertices then it will take 6 jumps to reach E.
∴ We will not find any path such that the frog takes 5 jumps and reaches E.
∴ a5 = 0
Similarly, we find that only for even values of n the frog can reach E.
2n − 1 is an odd number
∴ No route is possible with odd number of jumps.
Hence, option (a).
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