By Abraham P Hillman

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It is easily seen that the explicit factorization is a n & b n ' (a & b)(a n&1 % a n&2b % a n&3b 2 % ... % ab n&2 % b n&1). Example 3. Prove that n(n2 + 5) is an integral multiple of 6 for all integers n, that is, there is an integer u such that n(n2 + 5) = 6u. Proof: We begin by proving the desired result for all the integers greater than or equal to 0 by mathematical induction. When n = 0, n(n2 + 5) is 0. Since 0 ' 6@0 is a multiple of 6, the result holds for n = 0. We now assume it true for n = k, and seek to derive from this its truth for n = k + 1.

Thus we see that 3 k for k = 0, 1, 2, 3 is the number of ways of choosing a subset of k elements from a set S of 3 elements. Similarly, one can see that the number of ways of choosing k elements from a set of n elements is n . k 51 For example, the set {1, 2, 3, 4, 5} with 5 elements has 5 3 subsets having 3 elements. Since 5 3 ' 5@4@3 ' 10, 1@2@3 it is not too difficult to write out all ten of these subsets as {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}.

For the sequence c0, c1, ... of the previous problem, express c12 in the form aFu + Fv, where Fu and Fv are Fibonacci numbers, and conjecture a similar expression for cm. 27 * 43. In the sequence 1/5, 3/5, 4/5, 9/10, 19/20, 39/40, ... each succeeding term is the average of the previous term and 1. Thus: 3 1 1 4 1 3 9 1 4 ' % 1 , ' % 1 , ' % 1 , ÿ. 5 2 5 5 2 5 10 2 5 (a) Show that the twenty-first term is 1 & 1 . 5@218 (b) Express the nth term similarly. (c) Sum the first five hundred terms. * 44.

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