... satisfies, in the formf(n) − f(n +2)=0.Since f(0) = 1 and f(1) = 0, you have found that f(n) = 1, if n is even, and f(n) = 0, if n is odd, and you’re all finished. If, on the other hand, you get a recurrence ... the summand function F (n, k).Suppose, for a moment, that we are trying to do the sum f(n)=kF (n, k).Suppose that we execute Zeilberger’s algorithm, and we find a recurrence of theform (6.1.3) ... k and obtainJj=0aj(n)f(n + j)=0, (6.1.4)assuming, say, that G(n, k) has compact support in k for each n. Now there areseveral possible scenarios.It might happen that J = 1, i.e., that...