{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Discrete Mathematics with Graph Theory (3rd Edition) 211

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

This preview shows page 1. Sign up to view the full content.

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 =
This is the end of the preview. Sign up to access the rest of the document.

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 + . .. + ; ....
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern