Answer to Question #3955 in Abstract Algebra for adiba1390
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
A(2n-1) = 0
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nice and simple way of explanation.... thanx