1. Recurrence Relations: Selected Exercises

2. 10 (a)
A person deposits $1,000 in an account that yields 9% interest compounded annually.
a) Set up a recurrence relation for the amount in the account at the end of n years.

3. 10 (a) Solution Let an represent the amount after n years.
an = an an-1 = 1.09an-1
a0 = 1000.

4. 10 (b)
A person deposits $1,000 in an account that yields 9% interest compounded annually.
Find an explicit formula for the amount in the account at the end of n years.

5. 10 (b) Solution After 1 year, a1 = 1.09a0 = 1.09x1000 = 1000x1.091
After 2 years, a2 = 1.09a1
= 1.09(1000x(1.09)1)
= 1000x(1.09)2
After n years, an = 1000x(1.09)n
Since an is recursively defined, we prove the formula, for n ≥ 0, by mathematical induction
(The problem does not ask for proof).

6. 10 (b) Solution Basis n = 0: a0 = 1000 = 1000x(1.09)0 .
The 1st equality is the recurrence relation’s initial condition.
Show: an = 1000x1.09n  an+1 = 1000x1.09n+1.
Assume an = 1000x1.09n .
an+1 = 1.09an = 1.09 (1000x1.09n) = 1000x1.09n+1.
The 1st equality is from the definition of the recurrence relation.
The 2nd equality is from the induction hypothesis.

7. 10 (c)
A person deposits $1,000 in an account that yields 9% interest compounded annually.
How much money will the account contain after 100 years?

8. 10 (c) Solution
The account will contain a100 dollars after 100 years: a100 = 1000x = $5,529,041.
That is before taxes .
With 30% federal + 10% CA on interest earned, it becomes 1000x = $131,500.

9. 20 A country uses as currency:
coins with pesos values of 1, 2, 5, & 10 pesos
bills with pesos values of 5, 10, 20, 50, & 100.
Find a recurrence relation, an, for the # of payment sequences for n pesos.
E.g., a bill of 4 pesos could be paid with any of the following sequences:
1, 1, 1, 1
1,1, 2
1, 2, 1
2, 1, 1
2, 2

10. 20 Solution
For n pesos, our 1st (order matters) currency object can be a coin or a bill.
Sequences that start w/ a 1 peso coin are different from other sequences:
Use the sum principle to decompose this problem into disjoint sub-problems, based on which kind of currency object starts the sequence.
If the 1st currency object is a coin, it could be a:
1 peso coin, in which case we have an-1 ways to finish the bill
2 peso coin, in which case we have an-2 ways to finish the bill
5 peso coin, in which case we have an-5 ways to finish the bill
10 peso coin, in which case we have an-10 ways to finish the bill
If there were only coins, the recurrence relation would be
an = an-1 + an-2 + an-5 + an-10 with 10 initial conditions, a1 = 1, a2 = 2, a3 = 3, a4 = 5, a5 = 9, a6 = 15, a7 = 26, a8 = 44, a9 = 75, a10 = 125

11. 20 Solution continued But, we also can use bills.
If the 1st currency object is a bill, it could be a
5 peso, in which case we have an-5 ways to finish the bill
10 peso, in which case we have an-10 ways to finish the bill
20 peso, in which case we have an-20 ways to finish the bill
50 peso, in which case we have an-50 ways to finish the bill
100 peso, in which case we have an-100 ways to finish the bill
So, using both coins & bills, we have
an = an-1 + an-2 + an-5 + an-10 + an-5 + an-10 + an-20 + an-50 + an-100
= an-1 + an-2 + 2an-5 + 2an an-20 + an-50 + an-100 ,
with 100 initial conditions, which I will not produce.

12. 30 (a)
A string that contains only 0s, 1s, & 2s is called a ternary string.
Find a recurrence relation for the # of ternary strings of length n that do not contain 2 consecutive 0s.

13. 30 (a) Solution
We subtract the # of “bad” strings, bn, , from the # of ternary strings, 3n.
We use the sum principle to decompose the problem into disjoint sub-problems, depending on what digit starts the string:
Case the string starts with a 1: bn-1 ways to finish the string.
Case the string starts with a 2: bn-1 ways to finish the string.
Case the string starts with a 0:
Case the remaining string starts with a 0: 3n-2 ways to finish the string.
Case the remaining string starts with a 1: bn-2 ways to finish the string.
Case the remaining string starts with a 2: bn-2 ways to finish the string.
Summing, bn = 2bn-1 + 2bn-2 + 3n-2

14. 30 (b)
b) What are the initial conditions?

15. 30 (b) Solution
b0 = b1 = 0.
Why do we need 2 initial conditions?

16. 30 (c)
How many ternary strings of length 6 contain 2 consecutive 0s?

17. 30 (c) Solution The number of such strings is b6.
Using bn = 2bn-1 + 2bn-2 + 3n-2, we compute:
b0 = b1 = 0. (Initial conditions)
b2 = 2b1 + 2b = 1
b3 = 2b2 + 2b = 2x1 + 2x = 5
b4 = 2b3 + 2b = 2x5 + 2x = 21
b5 = 2b4 + 2b = 2x21 + 2x = 79
b6 = 2b5 + 2b = 2x79 + 2x = 281.

18. 40
Find a recurrence relation, en, for the # of bit strings of length n with an even # of 0s.

19. 40 Solution Strings are sequences: Order matters: There is a 1st bit.
Use the sum principle to decompose the problem into disjoint sub-problems, based on their 1st bit:
The strings with an even # of 0s that begin with 1: en-1
The strings with an even # of 0s that begin with 0: 2n-1 - en-1
Summing, en = en-1 + 2n-1 - en-1 = 2n-1
Does this answer suggest an alternate explanation?
Remember this question when we study binomial coefficients.

21. 49
The variation we consider begins with people numbered 1, …, n, standing around a circle.
In each stage, every 2nd person still alive is killed until only 1 survives.
We denote the number of the survivor by J(n).
Determine the value of J(n) for 1  n  16.

22. 49 Solution Put 5 people, named 1, 2, 3, 4, & 5, in a circle.
Starting with 1, kill every 2nd person until only 1 person is left.
The sequence of killings is:
So, J(5) = 3.
Continuing, for each value of n, results in the following table.

23. 49 Solution n
J(n) 1 9 3 2 10 5 11 7 4 12 13 6 14 15 8 16

24. 50
Use the values you found in Exercise 49 to conjecture a formula for J(n).
Hint: Write n = 2m + k, where m, k N & k < 2m .

25. 50 Solution n J(n) 1 = 20 + 0 1 9 = 23 + 1 3 2 = 21 + 0 10 = 23 + 2 5
3 =
11 = 7
4 =
12 = 9
5 =
13 = 11
6 =
14 = 13
7 =
15 = 15
8 =
16 =

26. 50 Solution continued n J(n) 1 = 20 + 0 1 = 2*0 + 1 9 = 23 + 1
3 = 2*1 + 1
2 =
10 =
5 = 2*2 + 1
3 =
11 =
7 = 2*3 + 1
4 =
12 =
9 = 2*4 + 1
5 =
13 =
11 = 2*5 + 1
6 =
14 =
13 = 2*6 + 1
7 =
15 =
15 = 2*7 + 1
8 =
16 =

27. 50 Solution continued
So, if n = 2m + k, where m, k N & k < 2m , then J(n) = 2k + 1.
Check this for J(17).