{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Discrete Mathematics with Graph Theory (3rd Edition) 119

# Discrete Mathematics with Graph Theory (3rd Edition) 119 -...

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

Section 5.1 117 using the induction hypothesis. Now the sum of the last three terms is since 2k + 1 < 2(k + 1). Hence, as desired. Thus, by the Principle of Mathematical Induction, the given inequality is true for all n ~ 2. (g) For n = 2, 112 + ;2 = 1 + ~ = ~ < ~ = 2 -~, so the inequality holds. Now assume that k ~ 2 1 1 1 1 1 and that - + - + - + ... + - < 2 - - We have 12 22 32 k 2 k' 1 1 1 1 12 + 22 + 3 2 + ... + (k + 1)2 ( 1 1 1 1) 1 = 12 + 22 + 3 2 + ... + k2 + (k+1)2 < (2 - ~) + (k: 1 )2 using the induction hypothesis 2 (k + 1)2 - k < - k(k + 1)2 k 2 + k + 1 k 2 + k 1 = 2 - k(k + 1)2 < 2 - k(k + 1)2 = 2 - k + 1 ' which is the desired inequality with n = k + 1. By the Principle of Mathematical Induction, we conclude that the inequality holds for all n ~ 2. (h) Wh
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 - 1 1 -1 1 N 1 1 -1 fl v'2 v'2 ( . en n -,L. ...i=1 Vi -v'1 + v'2 -+ v'2' ow + 72 -+ 2 >""2 +""2 smce .J2 ~ 1.4) and 4 + l' = 24 = .J2. Thus, 2::~=1 ~ > .Jii when n = 2. Now suppose that k ~ 2 and the given assertion is true for n = k; that is, suppose 2:::=1 ~ > .jk. We have to prove that the assertion is true for n = k + 1; that is, we have to prove that 2:::':;:-; ~ > v'k"+1, Now k+l 1 L:-i=1 Vi k 1 1 L:-+-== i=1 Vi v'k"+1 1 > .jk + ~ (by the induction hypothesis) yk+1 .jkv'k"+1 + 1 v'k"+1 Vk2+k+1 > v'k2+1 = k+1 =Jk+1 v'k"+1 v'k"+1 Jk + 1 as desired. By the Principle of Mathematical Induction, the given assertion is true for all n ~ 2. (i) [BBJ With n = 2, 2(2) - 1 = 3 ~ 2(2) - 2 =2, so the inequality holds. Now assume k ~ 2 and...
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