411 REFERENCE · Wikibooks
Solutions To Mathematics Textbooks/Proofs and Fundamentals/Chapter 6
Reference material by Wikibooks contributors. This retains its source identity and is not 411 news reporting.
Section 6.2
Section 6.3
6.3.1
Show the following equation holds for all n in the set of all counting numbers.
(I) 1+3+5+...+(2n-1)=n^2
First we prove the base case. Note that 1 = 1^2. Next, assume that for any integer k, equation (I) holds. Add (2(k+1)-1)=2k+1 to both sides of (I). This yields 1+3+5+...+(2k-1)+(2(k+1)-1)=k^2+2k+1. Note that k^2+2k+1 = (k+1)^2. Thus, 1+3+5+...+(2k-1)+(2(k+1)-1)=(k+1)^2. Therefore, if (I) holds for a given integer k, it also holds for k+1. Since it has been shown that (I) holds for k=1, by induction, (I) is necessarily true for all n in the set of all counting numbers.
(II)
6.3.2
6.3.3
6.3.4
6.3.5
6.3.6
6.3.7
6.3.8
6.3.9
6.3.10
6.3.11
trivial
6.3.12
6.3.13
6.3.14
6.3.15
6.3.16
6.3.17