key5sp00 - COT3100-01, Spring 2000 S. Lang Solution Key to...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
COT3100-01, Spring 2000 Assigned: 4/12/2000 S. Lang Solution Key to Assignment #5 (40 pts.) Due: 4/19 in class 1. Use induction on n 0 to prove the following summation identity: . 0 2 ) 2 ( ) 1 ( 2 ) 1 ( = + - = - n i n n n i i We use induction on n 0. (Basis Step) Consider n = 0. In this case, . 0 2 0 2 ) 0 2 0 ( 0 ) 1 ( RHS The . 0 0 ) 1 ( ) 1 ( LHS The 2 0 0 0 2 = = + - = = - = ∑ - = = i i i Thus, LHS = RHS, so the Basis Step is proved. (Induction Hypothesis) Consider n = k . Suppose . 0 some for 0 2 ) 2 ( ) 1 ( 2 ) 1 ( = + - = - k k i k k k i i (Induction Step) Consider n = k + 1. We need to prove (1). of RHS 2 ) 2 3 ( 1 ) 1 ( 2 ) 2 4 2 2 ( 1 ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( since , 2 ) 1 ( 2 ) 1 ( 2 ) 2 )( 1 ( 1 ) 1 ( Hypothesis Induction by the , ) 1 ( ) 1 ( 2 ) 2 ( ) 1 ( summation of definition by the , ) 1 ( ) 1 ( ) 1 ( (1) of LHS The 2 2 1 2 2 1 2 1 2 1 0 2 = + + + - = + + + - - + - = - - = - = - + - + + - + - = + - + + - = + - + ∑ - = + + + + + = k k k k k k k k k k k k k k k k k i k k k k k k k i i Thus, the Induction Step is proved. By induction, we proved that 2. Suppose a function f ( n ) is defined for integer n 0 recursively by the following recurrence: f (0) = 7; f (1) = 14; and f ( n ) = f ( n – 1) + 12 f ( n – 2) for n 2. Prove that f ( n ) = 5(4) n + 2(–3) n for all n 0. We use strong induction on
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 07/14/2011.

Page1 / 3

key5sp00 - COT3100-01, Spring 2000 S. Lang Solution Key to...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online