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Discrete Mathematics with Graph Theory (3rd Edition) 211

Discrete Mathematics with Graph Theory (3rd Edition) 211 -...

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Section 7.7 n(n - l)(n - 2) n(n - 1) = n + 2n(n - 1) + 3 + ... + (n - 1)2 + (n - 1)2n + n 2 2 2 = n[ 1 + 2(n -1) + 3 (n - 1)2(n - 2) + ... + (n _ 2)2 (n; 1) + (n _ 1)2 + n] =n[1+2(n~1) +3(n;1) + ... +(n_2)(n;1)1+(n_1)(n~1) +n] =n~(k+1)(n~1)=n[~k(n~1) + ~(n~l)] = n[(n _1)2 n - 2 + 2 n - 1 ] = n2n-2(n -1 + 2) = n(n + 1)2 n - 2 using (a) and the result of Problem 34 in the second last line. 209 23. The remarks at the beginning of this section suggest that the terms of the Fibonacci sequence are special kinds of diagonal sums. Using Figures 7.8 and 7.6, for instance, we see that 2 = a2 = @ + (D, 3 = a3 = @ + m, 5 = a4 = @ + m + @. We conjecture that where we sum over all k for which n - k ?: k. Using identity (5) and the strong form of the Principle of Mathematical Induction, we can verify this result. For n = 1, al = 1 = (~), while for n = 2, a2 =
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Unformatted text preview: + (D. Now assume that k > 2 and the result is true for all n in the range 1 ~ n < k. We wish to prove the result for k. Note that the identity we wish to prove has two different types of ~ final terms depending on whether n is even or odd. When n is odd, the final term in the sum is (n~l) 2 while, if n is even, the final term is (%). We establish the odd and even cases separately. 2 Case 1: n is odd. ( n) (n -1) (n -2) (!!±l) In this case, we wish to prove that an = 0 + 1 + 2 + . .. + n~l . (Note how identity (5) was employed at the very first step.) Now applying the induction hypothesis to n - 1, the term in the first set of brackets here is an-l and applying the induction hypothesis to n - 2, the other term is a n -2. Case 2: n is even. ( n) (n -1) (n -2) (!!) This time we wish to prove an = 0 + 1 + 2 + . .. + ; ....
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