MTH 18202
Quiz 1 Version 2
(1 pt) Name _______Solution Key ____________
************************************* Show Your Work! ************************************
1. (7 pt) Let ! A = {1, 2, 3, 4 } . For the relation ! R = {(1, 1), (2,2), (2, 3), (3, 2), (4, 4)} on A, determine which of
the properties reflexive, symmetric, transitive R possesses. Justify your answer!
Solution:
Reflexive: No! Since ! 3 ∈A but ( 3, 3) ∉R .
Symmetric: Yes! Since if ! ( a, b ) ∈R then either
(i) ! a = b where ! a = 1, 2, 4 in which case ! ( a, b ) = (b, a) = ( a, a ) ∈R , or
(ii) ! a = 2 and ! b = 3 in which case ! ( b, a ) = ( 3, 2 ) ∈R , or
(iii) ! a = 3 and ! b = 2 in which case ! ( b, a ) = ( 2, 3) ∈R
!!
Transitive: No! Since ! ( 3, 2 ) ∈R and ( 2, 3) ∈R but ( 3, 3) ∉R.
(7 pt) Let ! {an } be the sequence defined recursively by ! a1 = 1, ! a2 = 3 , and ! an = 2an−1 + 3an−2 if ! n ≥ 3 .
a) Find ! a3 and ! a4 .
b) What is an explicit formula for ! an ?
Solution:
Solution: ! an = 3n−1 ! !
! a3 = 2a2 + 3a1 = 2 ⋅ 3 + 3⋅1 = 9 = 32 , and
! a4 = 2a3 + 3a1 = 2 ⋅ 9 + 3⋅ 3 = 27 = 33
!!
3. (10 pt) Let ! {an } be the sequence defined recursively by ! a1 = 1 , ! a2 = 4 , and ! an = 2an−1 + 8an−2 if ! n ≥ 3 .
Use strong induction to prove that ! P(n) : an = 4 n−1 is an explicit formula for ! an every positive integer n.
Solution:
Base Step: First, verify that ! P (1) and ! P(2) are both true:
! P(1) : ! 1 = a1 = 41−1 = 4 0 = 1 is true, and ! P(2) : ! 4 = a2 = 4 2−1 = 41 = 4 is true.
Induction Step:
A. The Strong Induction Hypothesis: Assume for ! n = some k ≥ 1 that ! P(i) : ai = 4 i−1 for ! 1 ≤ i ≤ k .
B. The Induction Step Proof: Using the Strong Induction Hypothesis, prove ! P(n) is true for ! n = k + 1 ≥ 2 ;
that is, prove ! P(k + 1) : ak+1 = 4 (k+1)−1 = 4 k .
Proof: We can assume that ! k ≥ 2 since by the base step, ! P (1) and ! P(2) are both true. This means that we
can assume ! k + 1 ≥ 3 . Thus, by the “if ! n ≥ 3 ” case of the sequence’s recursive definition with ! n = k + 1 ,
! ak+1 = 2a( k+1)−1 + 8a( k+1)−2 = 2ak + 8ak−1 .
Now both ! k, k − 1 ≥ 1 ; hence, by using the Strong Induction Hypothesis twice,
!
ak+1 = 2ak + 8ak−1 = 2 ⋅ 4 k−1 + 8 ⋅ 5( k−1)−1 = 2 ⋅ 4 k−1 + 2 ⋅ 4 ⋅ 4 ( k−1)−1
= 2 ⋅ 4 k−1 + 2 ⋅ 4 k−1 = ( 2 + 2 ) 4 k−1 = 4 ⋅ 4 k−1 = 4 k
which is the required result as shown in (B). This by the Principle of Strong Induction, ! P(n) : an = 4 n−1 is
true for all ! n ≥ 1 .
!!