# 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)?

Expert's answer

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'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

ANSWER

A(2n-1) = 0

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'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

ANSWER

A(2n-1) = 0

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## Comments

Assignment Expert09.08.12, 14:18You're welcome. We are glad to be helpful.

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gautam08.08.12, 19:27nice and simple way of explanation.... thanx

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