# Adv Alegbra HW Solutions 7 - secting the region in Figure...

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

7 1.13 (i) Prove that 2 n > n 3 for all n 10. Solution. Base step .2 10 = 1024 > 10 3 = 1000. (Note that 2 9 = 512 < 9 3 = 729.) Inductive step Note that n 10 implies n 4. The inductive hypothesis is 2 n > n 3 ; multiplying both sides by 2 gives 2 n + 1 = 2 2 n > 2 n 3 = n 3 + n 3 n 3 + 4 n 2 = n 3 + 3 n 2 + n 2 > n 3 + 3 n 2 + 4 n = n 3 + 3 n 2 + 3 n + n n 3 + 3 n 2 + 3 n + 1 = ( n + 1 ) 3 . (ii) Prove that 2 n > n 4 for all n 17. Solution. Base step .2 17 = 131 , 072 > 17 4 = 83 , 521. (Note that 16 4 = ( 2 4 ) 4 = 2 16 .) Inductive step . Note that n 17 implies n 7. The inductive hypothesis is 2 n > n 4 ; multiplying both sides by 2 gives 2 n + 1 = 2 2 n > 2 n 4 = n 4 + n 4 n 4 + 5 n 3 n 4 + 4 n 3 + n 3 n 4 + 4 n 3 + 7 n 2 n 4 + 4 n 3 + 6 n 2 + n 2 n 4 + 4 n 3 + 6 n 2 + 5 n n 4 + 4 n 3 + 6 n 2 + 4 n + 1 = ( n + 1 ) 4 . 1.14 Around 1350, N. Oresme was able to sum the series n = 1 n / 2 n by dis-
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: secting the region in Figure 1.4 in two ways. Let A n be the vertical rect-angle with base 1 2 n and height n , so that area ( A n ) = n / 2 n , and let B n be horizontal rectangle with base 1 2 n + 1 2 n + 1 + ··· and height 1. Prove that ∑ ∞ n = 1 n / 2 n = 2. Solution. You may assume that ∑ ∞ n = ar n = a /( 1 − r ) if 0 ≤ r < 1. Now compute the area using ∑ A n = ∑ B n ....
View Full Document

## This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

Ask a homework question - tutors are online