# Answer to Question #3955 in Abstract Algebra for adiba1390

Question #3955
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 a(n) be the number of distinct paths of exactly n jumps ending in E. Then what is the value of a(2n – 1)?
Firstly, notice that the shortest path from A to E involves exactly 4 jumps.
4 is even number. We can prove that the frog always makes even number of
jumps to reach vertex E.
Assume we have some path from A to E. If at some vertex our path turns back
then later it must turn back again, so this vertex is visited twice.
Such way the evenness of length of the path can&#039;t be corrupted. Thus, any
possible path has even length.
Since (2n-1) is odd, there is no path from A to E that frog can follow for (2n-1) jumps.
Thus, a(2n-1) = 0
A(2n-1) = 0

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Assignment Expert
09.08.12, 14:18

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gautam
08.08.12, 19:27

nice and simple way of explanation.... thanx