Solving Recurrence Relations

Section 2.4 Solving Recurrence Relations

Investigate!

Consider the recurrence relation

a_{n} = 5 a_{n - 1} - 6 a_{n - 2} .

What sequence do you get if the initial conditions are $a_{0} = 1,$ $a_{1} = 2 ?$ Give a closed formula for this sequence.
What sequence do you get if the initial conditions are $a_{0} = 1,$ $a_{1} = 3 ?$ Give a closed formula.
What if $a_{0} = 2$ and $a_{1} = 5 ?$ Find a closed formula.

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We have seen that it is often easier to find recursive definitions than closed formulas. Lucky for us, there are a few techniques for converting recursive definitions to closed formulas. Doing so is called solving a recurrence relation. Recall that the recurrence relation is a recursive definition without the initial conditions. For example, the recurrence relation for the Fibonacci sequence is

F_{n} = F_{n - 1} + F_{n - 2} .

(This, together with the initial conditions

F_{0} = 0

and

F_{1} = 1

give the entire recursive definition for the sequence.)

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Example 2.4.1.

Find a recurrence relation and initial conditions for

1, 5, 17, 53, 161, 485 \dots .

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Solution.

Finding the recurrence relation would be easier if we had some context for the problem (like the Tower of Hanoi, for example). Alas, we have only the sequence. Remember, the recurrence relation tells you how to get from previous terms to future terms. What is going on here? We could look at the differences between terms:

4, 12, 36, 108, \dots .

Notice that these are growing by a factor of 3. Is the original sequence as well?

1 \cdot 3 = 3,

5 \cdot 3 = 15,

17 \cdot 3 = 51

and so on. It appears that we always end up with 2 less than the next term. Aha!

a_{n} = 3 a_{n - 1} + 2

is our recurrence relation and the initial condition is

a_{0} = 1 .

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We are going to try to solve these recurrence relations. By this we mean something very similar to solving differential equations: we want to find a function of

n

(a closed formula) which satisfies the recurrence relation, as well as the initial condition.

Recurrence relations are sometimes called difference equations since they can describe the difference between terms and this highlights the relation to differential equations further.

Just like for differential equations, finding a solution might be tricky, but checking that the solution is correct is easy.

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Example 2.4.2.

Check that

a_{n} = 2^{n} + 1

is a solution to the recurrence relation

a_{n} = 2 a_{n - 1} - 1

with

a_{1} = 3 .

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Solution.

First, it is easy to check the initial condition:

a_{1}

should be

2^{1} + 1

according to our closed formula. Indeed,

2^{1} + 1 = 3,

which is what we want. To check that our proposed solution satisfies the recurrence relation, try plugging it in.

\begin{aligned} 2 a_{n - 1} - 1 & = 2 (2^{n - 1} + 1) - 1 \\ = 2^{n} + 2 - 1 \\ = 2^{n} + 1 \\ = a_{n} . \end{aligned}

That’s what our recurrence relation says! We have a solution.

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Sometimes we can be clever and solve a recurrence relation by inspection. We generate the sequence using the recurrence relation and keep track of what we are doing so that we can see how to jump to finding just the

a_{n}

term. Here are two examples of how you might do that.

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Telescoping refers to the phenomenon when many terms in a large sum cancel out—so the sum “telescopes.” For example:

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(2 - 1) + (3 - 2) + (4 - 3) + \dots + (100 - 99) + (101 - 100) = - 1 + 101

because every third term looks like:

2 + - 2 = 0,

and then

3 + - 3 = 0

and so on.

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We can use this behavior to solve recurrence relations. Here is an example.

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Example 2.4.3.

Solve the recurrence relation

a_{n} = a_{n - 1} + n

with initial term

a_{0} = 4 .

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Solution.

To get a feel for the recurrence relation, write out the first few terms of the sequence:

4, 5, 7, 10, 14, 19, \dots .

Look at the difference between terms.

a_{1} - a_{0} = 1

and

a_{2} - a_{1} = 2

and so on. The key thing here is that the difference between terms is

n .

We can write this explicitly:

a_{n} - a_{n - 1} = n .

Of course, we could have arrived at this conclusion directly from the recurrence relation by subtracting

a_{n - 1}

from both sides.

Now use this equation over and over again, changing

n

each time:

\begin{aligned} a_{1} - a_{0} & = 1 \\ a_{2} - a_{1} & = 2 \\ a_{3} - a_{2} & = 3 \\ ⋮ & ⋮ \\ a_{n} - a_{n - 1} & = n . \end{aligned}

Add all these equations together. On the right-hand side, we get the sum

1 + 2 + 3 + \dots + n .

We already know this can be simplified to

\frac{n (n + 1)}{2} .

What happens on the left-hand side? We get

(a_{1} - a_{0}) + (a_{2} - a_{1}) + (a_{3} - a_{2}) + \dots (a_{n - 1} - a_{n - 2}) + (a_{n} - a_{n - 1}) .

This sum telescopes. We are left with only the

- a_{0}

from the first equation and the

a_{n}

from the last equation. Putting this all together we have

- a_{0} + a_{n} = \frac{n (n + 1)}{2}

a_{n} = \frac{n (n + 1)}{2} + a_{0} .

But we know that

a_{0} = 4 .

So the solution to the recurrence relation, subject to the initial condition is

a_{n} = \frac{n (n + 1)}{2} + 4 .

(Now that we know that, we should notice that the sequence is the result of adding 4 to each of the triangular numbers.)

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The above example shows a way to solve recurrence relations of the form

a_{n} = a_{n - 1} + f (n)

where

\sum_{k = 1}^{n} f (k)

has a known closed formula. If you rewrite the recurrence relation as

a_{n} - a_{n - 1} = f (n),

and then add up all the different equations with

n

ranging between 1 and

n,

the left-hand side will always give you

a_{n} - a_{0} .

The right-hand side will be

\sum_{k = 1}^{n} f (k),

which is why we need to know the closed formula for that sum.

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However, telescoping will not help us with a recursion such as

a_{n} = 3 a_{n - 1} + 2

since the left-hand side will not telescope. You will have

- 3 a_{n - 1}

’s but only one

a_{n - 1} .

However, we can still be clever if we use iteration.

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We have already seen an example of iteration when we found the closed formula for arithmetic and geometric sequences. The idea is, we iterate the process of finding the next term, starting with the known initial condition, up until we have

a_{n} .

Then we simplify. In the arithmetic sequence example, we simplified by multiplying

d

by the number of times we add it to

a

when we get to

a_{n},

to get from

a_{n} = a + d + d + d + \dots + d

a_{n} = a + d n .

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To see how this works, let’s go through the same example we used for telescoping, but this time use iteration.

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Example 2.4.4.

Use iteration to solve the recurrence relation

a_{n} = a_{n - 1} + n

with

a_{0} = 4 .

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Solution.

Again, start by writing down the recurrence relation when

n = 1 .

This time, don’t subtract the

a_{n - 1}

terms to the other side:

a_{1} = a_{0} + 1 .

Now

a_{2} = a_{1} + 2,

but we know what

a_{1}

is. By substitution, we get

a_{2} = (a_{0} + 1) + 2 .

Now go to

a_{3} = a_{2} + 3,

using our known value of

a_{2} :

a_{3} = ((a_{0} + 1) + 2) + 3 .

We notice a pattern. Each time, we take the previous term and add the current index. So

a_{n} = ((((a_{0} + 1) + 2) + 3) + \dots + n - 1) + n .

Regrouping terms, we notice that

a_{n}

is just

a_{0}

plus the sum of the integers from

1

n .

So, since

a_{0} = 4,

a_{n} = 4 + \frac{n (n + 1)}{2} .

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Of course in this case we still needed to know formula for the sum of

1, \dots, n .

Let’s try iteration with a sequence for which telescoping doesn’t work.

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Example 2.4.5.

Solve the recurrence relation

a_{n} = 3 a_{n - 1} + 2

subject to

a_{0} = 1 .

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Solution.

Again, we iterate the recurrence relation, building up to the index

n .

\begin{aligned} a_{1} & = 3 a_{0} + 2 \\ a_{2} & = 3 (a_{1}) + 2 = 3 (3 a_{0} + 2) + 2 \\ a_{3} & = 3 [a_{2}] + 2 = 3 [3 (3 a_{0} + 2) + 2] + 2 \\ ⋮ & ⋮ ⋮ \\ a_{n} & = 3 (a_{n - 1}) + 2 = 3 (3 (3 (3 \dots (3 a_{0} + 2) + 2) + 2) \dots + 2) + 2 . \end{aligned}

It is difficult to see what is happening here because we have to distribute all those 3’s. Let’s try again, this time simplifying a bit as we go.

\begin{aligned} a_{1} & = 3 a_{0} + 2 \\ a_{2} & = 3 (a_{1}) + 2 = 3 (3 a_{0} + 2) + 2 = 3^{2} a_{0} + 2 \cdot 3 + 2 \\ a_{3} & = 3 [a_{2}] + 2 = 3 [3^{2} a_{0} + 2 \cdot 3 + 2] + 2 = 3^{3} a_{0} + 2 \cdot 3^{2} + 2 \cdot 3 + 2 \\ ⋮ & ⋮ ⋮ \\ a_{n} & = 3 (a_{n - 1}) + 2 = 3 (3^{n - 1} a_{0} + 2 \cdot 3^{n - 2} + \dots + 2) + 2 \\ = 3^{n} a_{0} + 2 \cdot 3^{n - 1} + 2 \cdot 3^{n - 2} + \dots + 2 \cdot 3 + 2 . \end{aligned}

Now we simplify.

a_{0} = 1,

so we have

3^{n} + ⟨ stuff ⟩ .

Note that all the other terms have a 2 in them. In fact, we have a geometric sum with first term

2

and common ratio

3 .

We have seen how to simplify

2 + 2 \cdot 3 + 2 \cdot 3^{2} + \dots + 2 \cdot 3^{n - 1} .

We get

\frac{2 - 2 \cdot 3^{n}}{- 2}

which simplifies to

3^{n} - 1 .

Putting this together with the first

3^{n}

term gives our closed formula:

a_{n} = 2 \cdot 3^{n} - 1 .

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Iteration can be messy, but when the recurrence relation only refers to one previous term (and maybe some function of

n

) it can work well. However, trying to iterate a recurrence relation such as

a_{n} = 2 a_{n - 1} + 3 a_{n - 2}

will be way too complicated. We would need to keep track of two sets of previous terms, each of which were expressed by two previous terms, and so on. The length of the formula would grow exponentially (double each time, in fact). Luckily there happens to be a method for solving recurrence relations which works very well on relations like this.

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Subsection The Characteristic Root Technique

Suppose we want to solve a recurrence relation expressed as a combination of the two previous terms, such as

a_{n} = a_{n - 1} + 6 a_{n - 2} .

In other words, we want to find a function of

n

which satisfies

a_{n} - a_{n - 1} - 6 a_{n - 2} = 0 .

Now iteration is too complicated, but think just for a second what would happen if we did iterate. In each step, we would, among other things, multiply a previous iteration by 6. So our closed formula would include

6

multiplied some number of times. Thus it is reasonable to guess the solution will contain parts that look geometric. Perhaps the solution will take the form

r^{n}

for some constant

r .

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The nice thing is, we know how to check whether a formula is actually a solution to a recurrence relation: plug it in. What happens if we plug in

r^{n}

into the recursion above? We get

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r^{n} - r^{n - 1} - 6 r^{n - 2} = 0 .

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Now solve for

r :

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r^{n - 2} (r^{2} - r - 6) = 0,

so by factoring,

r = - 2

r = 3

(or

r = 0,

although this does not help us). This tells us that

a_{n} = (- 2)^{n}

is a solution to the recurrence relation, as is

a_{n} = 3^{n} .

Which one is correct? They both are, unless we specify initial conditions. Notice we could also have

a_{n} = (- 2)^{n} + 3^{n} .

a_{n} = 7 (- 2)^{n} + 4 \cdot 3^{n} .

In fact, for any

a

and

b,

a_{n} = a (- 2)^{n} + b 3^{n}

is a solution (try plugging this into the recurrence relation). To find the values of

a

and

b,

use the initial conditions.

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This points us in the direction of a more general technique for solving recurrence relations. Notice we will always be able to factor out the

r^{n - 2}

as we did above. So we really only care about the other part. We call this other part the characteristic equation for the recurrence relation. We are interested in finding the roots of the characteristic equation, which are called (surprise) the characteristic roots.

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Characteristic Roots.

Given a recurrence relation

a_{n} + α a_{n - 1} + β a_{n - 2} = 0,

the characteristic polynomial is

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x^{2} + α x + β

giving the characteristic equation:

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x^{2} + α x + β = 0 .

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r_{1}

and

r_{2}

are two distinct roots of the characteristic polynomial (i.e., solutions to the characteristic equation), then the solution to the recurrence relation is

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a_{n} = a r_{1}^{n} + b r_{2}^{n},

where

a

and

b

are constants determined by the initial conditions.

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Example 2.4.6.

Solve the recurrence relation

a_{n} = 7 a_{n - 1} - 10 a_{n - 2}

with

a_{0} = 2

and

a_{1} = 3 .

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Solution.

Rewrite the recurrence relation

a_{n} - 7 a_{n - 1} + 10 a_{n - 2} = 0 .

Now form the characteristic equation:

x^{2} - 7 x + 10 = 0

and solve for

x :

(x - 2) (x - 5) = 0

x = 2

and

x = 5

are the characteristic roots. We therefore know that the solution to the recurrence relation will have the form

a_{n} = a 2^{n} + b 5^{n} .

To find

a

and

b,

plug in

n = 0

and

n = 1

to get a system of two equations with two unknowns:

\begin{aligned} 2 & = a 2^{0} + b 5^{0} = a + b \\ 3 & = a 2^{1} + b 5^{1} = 2 a + 5 b \end{aligned}

Solving this system gives

a = \frac{7}{3}

and

b = - \frac{1}{3}

so the solution to the recurrence relation is

a_{n} = \frac{7}{3} 2^{n} - \frac{1}{3} 5^{n} .

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Perhaps the most famous recurrence relation is

F_{n} = F_{n - 1} + F_{n - 2},

which together with the initial conditions

F_{0} = 0

and

F_{1} = 1

defines the Fibonacci sequence. But notice that this is precisely the type of recurrence relation on which we can use the characteristic root technique. When you do, the only thing that changes is that the characteristic equation does not factor, so you need to use the quadratic formula to find the characteristic roots. In fact, doing so gives the third most famous irrational number,

φ,

the golden ratio.

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Before leaving the characteristic root technique, we should think about what might happen when you solve the characteristic equation. We have an example above in which the characteristic polynomial has two distinct roots. These roots can be integers, or perhaps irrational numbers (requiring the quadratic formula to find them). In these cases, we know what the solution to the recurrence relation looks like.

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However, it is possible for the characteristic polynomial to have only one root. This can happen if the characteristic polynomial factors as

(x - r)^{2} .

It is still the case that

r^{n}

would be a solution to the recurrence relation, but we won’t be able to find solutions for all initial conditions using the general form

a_{n} = a r_{1}^{n} + b r_{2}^{n},

since we can’t distinguish between

r_{1}^{n}

and

r_{2}^{n} .

We are in luck though:

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Characteristic Root Technique for Repeated Roots.

Suppose the recurrence relation

a_{n} = α a_{n - 1} + β a_{n - 2}

has a characteristic polynomial with only one root

r .

Then the solution to the recurrence relation is

🔗

a_{n} = a r^{n} + b n r^{n}

where

a

and

b

are constants determined by the initial conditions.

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Notice the extra

n

b n r^{n} .

This allows us to solve for the constants

a

and

b

from the initial conditions.

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Example 2.4.7.

Solve the recurrence relation

a_{n} = 6 a_{n - 1} - 9 a_{n - 2}

with initial conditions

a_{0} = 1

and

a_{1} = 4 .

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Solution.

The characteristic polynomial is

x^{2} - 6 x + 9 .

We solve the characteristic equation

x^{2} - 6 x + 9 = 0

by factoring:

(x - 3)^{2} = 0

x = 3

is the only characteristic root. Therefore we know that the solution to the recurrence relation has the form

a_{n} = a 3^{n} + b n 3^{n}

for some constants

a

and

b .

Now use the initial conditions:

\begin{aligned} a_{0} = 1 & = a 3^{0} + b \cdot 0 \cdot 3^{0} = a \\ a_{1} = 4 & = a \cdot 3 + b \cdot 1 \cdot 3 = 3 a + 3 b . \end{aligned}

Since

a = 1,

we find that

b = \frac{1}{3} .

Therefore the solution to the recurrence relation is

a_{n} = 3^{n} + \frac{1}{3} n 3^{n} .

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Although we will not consider examples more complicated than these, this characteristic root technique can be applied to much more complicated recurrence relations. For example,

a_{n} = 2 a_{n - 1} + a_{n - 2} - 3 a_{n - 3}

has characteristic polynomial

x^{3} - 2 x^{2} - x + 3 .

Assuming you see how to factor such a degree 3 (or more) polynomial you can easily find the characteristic roots and as such solve the recurrence relation (the solution would look like

a_{n} = a r_{1}^{n} + b r_{2}^{n} + c r_{3}^{n}

if there were 3 distinct roots). It is also possible that the characteristics roots are complex numbers.

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However, the characteristic root technique is only useful for solving recurrence relations in a particular form:

a_{n}

is given as a linear combination of some number of previous terms. These recurrence relations are called linear homogeneous recurrence relations with constant coefficients. The “homogeneous” refers to the fact that there is no additional term in the recurrence relation other than a multiple of

a_{j}

terms. For example,

a_{n} = 2 a_{n - 1} + 1

is non-homogeneous because of the additional constant 1. There are general methods of solving such things, but we will not consider them here, other than through the use of telescoping or iteration described above.

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Exercises Exercises

1.

Find the next two terms in

(a_{n})_{n \geq 0}

beginning

3, 5, 11, 21, 43, 85 \dots .

Then give a recursive definition for the sequence. Finally, use the characteristic root technique to find a closed formula for the sequence.

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2.

Consider the sequences

2, 5, 12, 29, 70, 169, 408, \dots

(with

a_{0} = 2

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Describe the rate of growth of this sequence.
Find a recursive definition for the sequence.
Find a closed formula for the sequence.
If you look at the sequence of differences between terms, and then the sequence of second differences, the sequence of third differences, and so on, will you ever get a constant sequence? Explain how you know.

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3.

Solve the recurrence relation

a_{n} = a_{n - 1} + 2^{n}

with

a_{0} = 2 .

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a_{n} =

🔗

4.

Show that

4^{n}

is a solution to the recurrence relation

a_{n} = 3 a_{n - 1} + 4 a_{n - 2} .

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5.

Find the solution to the recurrence relation

a_{n} = - a_{n - 1} + 6 a_{n - 2}

with initial terms

a_{0} = 2

and

a_{1} = - 1 .

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a_{n} =

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6.

Find the solution to the recurrence relation

a_{n} = 5 a_{n - 1} + 14 a_{n - 2}

with initial terms

a_{0} = 2

and

a_{1} = 5 .

🔗

a_{n} =

🔗

Find the solution to the recurrence relation

b_{n} = 5 b_{n - 1} + 14 b_{n - 2}

with initial terms

b_{0} = 2

and

b_{1} = 11 .

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b_{n} =

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7.

Find the solution to the recurrence relation

a_{n} = 5 a_{n - 1} + 6 a_{n - 2}

with initial terms

a_{0} = 3

and

a_{1} = 13 .

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a_{n} =

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8.

Suppose that

r^{n}

and

q^{n}

are both solutions to a recurrence relation of the form

a_{n} = α a_{n - 1} + β a_{n - 2} .

Prove that

c \cdot r^{n} + d \cdot q^{n}

is also a solution to the recurrence relation, for any constants

c, d .

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9.

Think back to the magical candy machine at your neighborhood grocery store. Suppose that the first time a quarter is put into the machine 1 Skittle comes out. The second time, 4 Skittles, the third time 16 Skittles, the fourth time 64 Skittles, etc.

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Find both a recursive and closed formula for how many Skittles the nth customer gets.
Check your solution for the closed formula by solving the recurrence relation using the Characteristic Root technique.

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10.

Let

a_{n}

be the number of

1 \times n

tile designs you can make using

1 \times 1

squares available in 4 colors and

1 \times 2

dominoes available in 5 colors.

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First, find a recurrence relation to describe the problem. Explain why the recurrence relation is correct (in the context of the problem).
Write out the first 6 terms of the sequence $a_{1}, a_{2}, \dots .$
Solve the recurrence relation. That is, find a closed formula for $a_{n} .$

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11.

You have access to

1 \times 1

tiles which come in 2 different colors and

1 \times 2

tiles which come in 3 different colors. We want to figure out how many different

1 \times n

path designs we can make out of these tiles.

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Find a recursive definition for the sequence $a_{n}$ of paths of length $n .$
Solve the recurrence relation using the Characteristic Root technique.

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12.

Solve the recurrence relation

a_{n} = 2 a_{n - 1} - a_{n - 2} .

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What is the solution if the initial terms are $a_{0} = 1$ and $a_{1} = 2 ?$
What do the initial terms need to be in order for $a_{9} = 30 ?$
For which $x$ are there initial terms which make $a_{9} = x ?$

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13.

Consider the recurrence relation

a_{n} = 4 a_{n - 1} - 4 a_{n - 2} .

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Find the general solution to the recurrence relation (beware the repeated root).
Find the solution when $a_{0} = 1$ and $a_{1} = 2 .$
Find the solution when $a_{0} = 1$ and $a_{1} = 8 .$

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