CSCE 235
Homework
Assignment 9 – Solution
Assigned: April 7, 2008
Due: 12:30 p.m. April 11,
2008
(Homework 5 minutes late will not be
accepted)
Recursively Defined Sequences
1. (5
points) Give recursive definitions of
each of the following sequences. For
each, give the values of the first two terms of the sequence as 
and 
; then define 
where k can
be 0, 1, or 2; and finally 
.
(a) 5, 3, 1, -1, -3, …
(b) 4, 1, 3, -2, 5, -7, 12, -19, 31, …
(c) 1, 2, 0, 3, -1, 4, -2, …
Solution
(a)
Given the sequence (5, 3, 1, -1, -3, …), = 5, = 3. The difference between a term and its
previous term is always 2! Thus, 
for
.
(b) Given the sequence (4, 1, 3, -2, 5, -7, 12,
-19, 31, …), = 4, = 1. After looking at the sequence, we cannot come
up with a formula that depends only on one previous term easily. So, we look at two previous terms. We see that the current term is always the
difference between the immediately previous two terms. Thus, 
for
.
(c)
Given the sequence (1, 2, 0, 3, -1, 4, -2, …), = 1, = 2. Now, this is a bit tricky. If you look at the even positions, you have
(1, 0, -1, -2, …).
If you look at the odd positions, you have (2, 3, 4,
…). So, we kind of know that 
for the
even-positioned terms; and 
. We can combine the
two formulas together using the fact that 
alternates as –1 and
1! Thus, we have 
for .
2. (10 points) Consider a sequence defined by 
, 
for . Prove by
induction that the sequence can be expressed as 
.
Solution
First, we will
establish the basis. When n = 2, = 
= 4-1 = 3 = 
= 
= 3. Thus, by
inspection, we have shown that 
is true for n =
2.
Next, we will establish the induction step.
So, suppose that for some n = k. Thus, we have: 
. Now, we need to show
that it is true also for some n = k + 1:
We need to show that 
.
We know that 
by definition. So, we have = 
= 
= 
= 
= 
= 
.
Thus, we have
shown that is true.
Thus, by the
Principle of Mathematical Induction, we have shown that the sequence defined by
can be expressed as .
3. (15 points) A sequence is defined recursively by 
, 
, and 
for .
(a) Find the first five terms of this sequence.
(b)
Guess a formula for 
.
(c) Prove by induction that your guess in (b) is correct.
Solution
(a) , , 
, 
, 
, 
, …
(b) Now,
we can guess a formula for . (Note that we can
actually find this out using the recursion relation theorem. But let us
guess!) The sequence is (2, 3, 5, 9, 17,
33, …) We see
that 3 is 2+1, 5 is 4+1, 9 is 8+1, 17 is 16+1, 33 is 32+1, … Thus, we can guess that 
is involved somehow. So, a first guess is 
. When
, 
, and so on. So, our
guess is valid. Thus, .
(c) Now,
we need to prove by induction that our guess is correct. First, we want to establish the basis: When 
, 
. So, the basis is
true by inspection.
Next,
we want to establish the induction step.
We know that for some n. Now, we need to show that 
. So, we know that 
by the definition of
the sequence. We know that
. What about 
? Can we say 
? If we use the strong
form of the Principle of Mathematical Induction, then yes. Substituting those two into the equation, we
have = 
= 
=
= 
=
. Thus, we have shown
that is true.
Therefore, by the Principle of
Mathematical Induction, we show that is correct for the sequence , for .
Recursion Relations
4. (70 points) In each of the following cases, give
an explicit formula for .
(a) 
, 
, and 
for .
(b) , 
for .
(c) 
, 
, and 
for .
(d) 
, 
, and 
for . Here c
and d are unspecified constants.
(e) , 
, and 
for .
(f) 
, 
, and 
for .
(g) , 
, and 
for .
Solution
(a) , , and for .
In
this case, a = -1, b = 6. The
characteristic equation that we need to solve is 
; that is: 
. Solving for r
gives us 
and we have two
characteristic roots: 
or 
. So, we know that we
can express in terms of 
, or 
. Now, we can solve
for the constants 
and 
.
, 
.
, 
.
, .
So, we know that 
, and, thus, 
. Thus, the solution
is = 
.
(b) , for .
In
this case, a = 5, b = 0. The
characteristic equation that we need to solve is ; that is: 
. Solving for r
gives us 
and we have one
characteristic root! So, we know that we
can express in terms of 
, or 
. Now, we can solve
for the constants and . 
. So, know that 
. We know that 
. Thus, we have 
. Thus, 
. The solution is 
.
(c) , , and for .
In this case,
a = 4, b = -4. The characteristic
equation that we need to solve is ; that is: 
. Solving for r
gives us 
and we have one
characteristic root: . So, we know that we
can express in terms of , or 
. Now, we can solve
for the constants and . 
. So, know that . Also, we have 
. We have 
. The solution is 
.
(d) , , and for . Here c
and d are unspecified constants.
In
this case, a = 5, b = -6. The
characteristic equation that we need to solve is ; that is: 
. Solving for r
gives us 
and we have two
characteristic roots: or 
! So, we know that we
can express in terms of , or 
. Now, we can solve
for the constants and .
, 
, 
So,
we know that 
, and 
. Thus, the solution is 
.
(e) , , and for .
In
this case, a = 0, b = 1. The
characteristic equation that we need to solve is ; that is: 
. Solving for r
gives us 
or 
. So we have two
characteristic roots! So, we know that
we can express in terms of , or 
. Now, we can solve
for the constants and .
, 
So, we know that 
, and 
. Thus, the solution is 
.
(f) , , and for .
In
this case, a = 0, b = 3. The
characteristic equation that we need to solve is ; that is: 
. Solving for r
gives us 
or 
. So we have two
characteristic roots! So, we know that
we can express in terms of , or 
. Now, we can solve
for the constants and .
, 
, 
,
So, we know that 
, and 
. Thus, the solution is 
.
(g) , , and for .
In
this case, a = -2, b = 3. The
characteristic equation that we need to solve is ; that is: 
. Solving for r
gives us 
. So we have two
characteristic roots: or ! So, we know that we
can express in terms of 
, or 
. Now, we can solve
for the constants and .
, 
So,
we know that , and . Thus, the solution is 
.
Problems
based on (Goodaire and Parmenter 2002) and (Ross and Wright 1988).