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Discrete Mathematics with Graph Theory (3rd Edition) 164

# Discrete Mathematics with Graph Theory (3rd Edition) 164 -...

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162 provethatfk < G)k-l. Now fk = fk-l + ik-2 < (D(k-l)-l + G)(k-2)-l = (D k - 3 G + 1) = G)k-3 (11) < G)k-3 (f~) = G)k-l Solutions to Review Exercises (applying the induction hypothesis to k - 1 and to k - 2, both of which are in the range 2 ~ e < k where the result holds) as required. By the Principle of Mathematical Induction (strong form), we conclude that the result holds for all n :::: l. 21. For n = 1,12 = 1 = 2 - 1 = h - 1, so the formula holds. Now assume k :::: 1 and that the formula holds for n = k. We have 12 +!4 + ... + f2(k+l) = 12 + f4 + ... + hk+2 = (12 + f4 + ... + 12k) + hk+2 = (f2k+l - 1) + hk+2 = hk+3- 1 by the induction hypothesis using the definition of hk+3 and this is the desired formula with n = k + 1 since 2(k + 1) + 1 = 2k + 3. By the Principle of Mathematical Induction, the formula is correct for all n :::: 1. 22. Let ai be the number of ways in which you can reach stair i. Then al = 1, a2 = 2 (one single step or one double step) and, in general, you can reach stair n in one step from stair n - 1 or a double step from stair
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