• Title: "Discrete Mathematics and Its Applications" Third Edition
• Author: Kenneth H. Rosen

P315:
3. Is the sequence {a_n} a solution of the recurrence relation a_n = 8a_n-1 - 16a_n-2 if
a) a_n = 0?
Answer: Yes.
      a_n = 0, a_n-1 = 0, a_n-2 = 0
=> 8a_n-1 - 16a_n-2 = 0 = a_n

c) a_n = 2ⁿ?
Answer: No.
      a_n = 2ⁿ, a_n-1 = 2^n-1, a_n-2 = 2^n-2
=> 8a_n-1 - 16a_n-2 = 8*2^n-1 - 16*2^n-2 = 4*2ⁿ - 4*2ⁿ = 0 != a_n

e) a_n = n4ⁿ?
Answer: Yes.
      a_n = n4ⁿ, a_n-1 = (n-1)*4^n-1, a_n-2 = (n-2)*4^n-2
=> 8a_n-1 - 16a_n-2 = 8*(n-1)*4^n-1 - 16*(n-2)*4^n-2
                                  = 2*(n-1)*4ⁿ - (n-2)*4ⁿ
                                  = [ 2(n-1) - (n-2) ] *4ⁿ
                                  = n4ⁿ
                                  = a_n
 

h) a_n = n²4ⁿ?
Answer: No.
       a_n = n²4ⁿ, a_n-1 = (n-1)²*4^n-1, a_n-2 = (n-2)²*4^n-2
=> 8a_n-1 - 16a_n-2 = 8*(n-1)²*4^n-1 - 16*(n-2)²*4^n-2
                                  = 2*(n-1)²*4ⁿ - (n-2)²*4ⁿ
                                  = [ 2(n-1)² - (n-2)² ] *4ⁿ
                                  = ( 2n² - 4n + 2 - n² + 4n - 4) *4ⁿ
                                  = (n² - 2)*4ⁿ
                                  != a_n
 
5. Find the solution to each of the following recurrence relations and initial conditions. Use an iterative approach such as that used in Example 5.
a) a_n = 3a_n-1 , a₀  = 2
Answer: a_n = 2 * 3ⁿ

c) a_n = a_n-1 + n, a₀  = 1
Answer: a_n = 1 + (n²+n)/2
a_n = a_n-1 + n
     = (a_n-2 + n-1) + n = a_n-2 + 2n - 1
     = (a_n-3 + n - 2 ) + 2n - 1=  a_n-3 + 3n - (1+2)
     = (a_n-4 + n - 3 ) + 3n - 3=  a_n-4 + 4n - (1+2+3)
     .
     .
     .
     = a_n-n + n*n - [1+2+3+ ... +(n-1)]
     = a₀ + n² - (n-1)(1+n-1)/2
     = 1 + n² - (n²-n)/2
     = 1 + (n²+n)/2

e) a_n = 2a_n-1 - 1, a₀  = 1
Answer: a_n  = 1
a_n = 2a_n-1 - 1
     = 2(2a_n-2 -1) - 1 = 2²a_n-2 - (1+2¹)
     = 2²(2a_n-3 - 1) - (2⁰ +2¹)  =  2³a_n-3 - (1+2¹+2²)
     = 2³(2a_n-4 - 1) - (2⁰+2¹+2²) =  2⁴a_n-4 - (1+2¹+2²+2³)
     .
     .
     .
     = 2ⁿa_n-n - (1+2¹+2²+ ... +2^n-1)
     = 2ⁿa₀ - (1+2¹+2²+ ... +2^n-1)
     = 2ⁿ * 1 - (2^n-1+1-1)/(2-1)
     = 2ⁿ - (2ⁿ-1)
     = 1
 
g) a_n = na_n-1, a₀  = 5
Answer: a_n  = 5 * n!
a_n = na_n-1
     = n[(n-1)a_n-2] = n(n-1)a_n-2
     = n(n-1)[(n-2)a_n-3] =  n(n-1)(n-2)a_n-3
     .
     .
     .
     =  n(n-1)(n-2) ... 1 * a_n-n
     = n! * a₀
     = 5 * n!
 

7. Suppose that the number of bacteria in a colony triples every hour.
a) Set up a recurrence relation for the number of bacteria after n hours have elapsed.
Answer: a_n = 3a_n-1, where a_n is the number of bacteria after n hours, and a₀ is the number of bacteria used to begin a new colony.
 
b) If 100 bacteria are used to begin a new colony, how many bacteria will be in the colony in 10 hours?
Answer:  5,904,900
Repeat the same steps as that in the problem 5, part a,
a_n = 3a_n-1
     = 3*(3a_n-2) = 3² a_n-2
     = 3² * 3a_n-3 = 3³ a_n-3
     .
     .
     .
     = 3ⁿ a_n-n
     = 3ⁿ a₀
Where a₀ = 100, so a₁₀ = 3¹⁰ a₀ = 3¹⁰ * 100 = 5,904,900
 
 

8. Assume that the population of the world in 1995 is 7 billion and is growing 3% a year.
a) Set up a recurrance relation for the population of the world n years after 1995.
Answer: a_n = a_n-1 + 3%a_n-1 = 1.03 * a_n-1, where a_n is the population of the world n years after 1995, and a₀ is the  population of the world in 1995, which is 7 billion.

b) Find an exiplicit formula for the population of the world n years after 1995.
Answer:  a_n = 7 * 1.03ⁿ billion
Repeat the similary steps to that in the problem 5, part a,
a_n = 1.03a_n-1
     = 1.03*(1.03a_n-2) = 1.03² a_n-2
     = 1.03² * 1.03a_n-3 = 1.03³ a_n-3
     .
     .
     .
     = 1.03ⁿ a_n-n
     = 1.03ⁿ a₀
     = 7 * 1.03ⁿ
Where a₀ = 7 billion.

c) What will be the population of the world be in 2010?
Answer: 10.9 billion
2010 - 1995 = 15
a₁₅ = 7 * 1.03¹⁵ = 10.9 billion
 
 

11. Use mathematical induction to verify the formula derived in Example 5 for the number of moves required to complete the Tower of Hanoi puzzle.
Proof: We are going to prove H_n = 2ⁿ - 1, the recurrence relation is H_n = 2H_n-1 + 1 with H₁ = 1.

• Basis step: H₁ = 2¹ - 1 = 1 => P(1) is true.
• Inductive step: Assume P(n) is true, i.e. H_n = 2ⁿ - 1

Then H_n+1 = 2H_n + 1
 = 2*(2ⁿ - 1) + 1
 = 2ⁿ⁺¹ - 2 + 1
 = 2ⁿ⁺¹ - 1
The last equation shows that P(n+1) is true. This completes the inductive step and completes the proof.

30. A bus driver pays all tolls, using only nickels and dimes, by throwing one coin at a time into the mechanical toll collector.
a) Find a recurrence relation for the number of different ways the bus driver can pay a toll of n cents (where the order in which the coins are used matters).
Answer: Let a_n be the number of ways the bus driver can pay a toll of n cents by using nickels and dimes ( where the order in which the coins are used matters).

1. If the toll can be devided by 5, i.e. n = 5k ( where k is a nonnegative integer ), then the number of ways the bus driver can pay a toll of n = 5k cents by using nickels and dimes is a_5k.
2. If the toll can't be devided by 5, i.e. n = 5k+1, 5k+2, 5k+3 or 5k+4 (where k is a nonnegative integer), then the number of ways the bus driver can pay n = 5k+1, 5k+2, 5k+3 or 5k+4 cents is a_5(k+1).

• if the last coin is a nickel, then we need to find all ways of paying 5(k-1), which is a_k-1.
• If the last coin is a dime, it is preceeded by all the possible ways of paying 5(k-2) which is a_k-2.

b) In how many different ways can the driver pay a toll of 45 cents?
Answer: 55 ways.
       a_k = a_k-1 + a_k-2, where a₀ = 1, a₁ = 1
      a₀ = 1,
      a₁ = 1,
      a₂ = a₀ + a₁ = 1+1 = 2
      a₃ = a₁ + a₂ = 1+2 = 3
      a₄ = a₂ + a₃ = 2+3 = 5
      a₅ = a₃ + a₄ = 3+5 = 8
      a₆ = a₄ + a₅ = 5+8 = 13
      a₇ = a₅ + a₆ = 8+13 = 21
      a₈ = a₆ + a₇ = 13+21 = 34
      a₉ = a₇ + a₈ = 21+34 = 55 ways
 
 

P324:
3. Solve the following recurrence relations together with the initial conditions given.
b) a_n = a_n-1 for n>=1, a₀ = 2
Answer: Obvious result: a_n = 2
OR this is a linear homogeneous recurrence relation with degree 1. We can do it like this:
      c₁ = 1,
      r - 1 = 0
=> r = 1,
      with the initial condition: a₀ = 2
=> a_n = 2

d) a_n = 4a_n-1 - 4a_n-2 for n>=2, a₀ = 6, a₁ = 8
Answer: a_n = 6 * 2ⁿ - 2n * 2ⁿ
      c₁ = 4, c₂ = -4,
      r² - 4r + 4 = 0
=> r₁ = r₂ = 2
=> a_n = b₁ * 2ⁿ + b₂ * n * 2ⁿ
      with the initial condition: a₀ = 6, a₁ = 8
=> 6 = b₁ * 2⁰ + b₂ * 0 * 2⁰ = b₁
      8 = b₁ * 2¹ + b₂ * 1 * 2¹ = 2b₁ + 2b₂
=> b₁ = 6, b₂ = -2
=> a_n = 6 * 2ⁿ - 2n * 2ⁿ

f) a_n = 4a_n-2 for n>=2, a₀ = 0, a₁ = 4
Answer: a_n = 2ⁿ - (-2)ⁿ
      c₁ = 0, c₂ = 4,
      r² - 4 = 0
=> r₁ = 2, r₂ = -2
=> a_n = b₁ * 2ⁿ + b₂ * (-2)ⁿ
      with the initial condition: a₀ = 0, a₁ = 4
=> 0 = b₁ * 2⁰ + b₂ * (-2)⁰ = b₁ + b₂
      4 = b₁ * 2¹ + b₂ * (-2)¹ = 2b₁ - 2b₂
=> b₁ = 1, b₂ = -1
=> a_n = 2ⁿ - (-2)ⁿ
 

 
4. Solve the following recurrence relations together with the initial conditions given.
a) a_n = a_n-1 + 6a_n-2 for n>=2, a₀ = 3, a₁ = 6
Answer: a_n = 3ⁿ * 12 / 5 + (-2)ⁿ * 3 / 5
      c₁ = 1, c₂ = 6,
      r² - r - 6 = 0
=> r₁ = 3, r₂ = -2
=> a_n = b₁ * 3ⁿ + b₂ * (-2)ⁿ
      with the initial condition: a₀ = 3, a₁ = 6
=> 3 = b₁ * 3⁰ + b₂ * (-2)⁰ = b₁ + b₂
      6 = b₁ * 3¹ + b₂ * (-2)¹ = 3b₁ - 2b₂
=> b₁ = 12/5, b₂ = 3/5
=> a_n = 3ⁿ * 12 / 5 + (-2)ⁿ * 3 / 5
 

c) a_n = 6a_n-1 - 8a_n-2 for n>=2, a₀ = 4, a₁ = 10
Answer: a_n = 3 * 2ⁿ + 4ⁿ
      c₁ = 6, c₂ = -8,
      r² - 6r + 8 = 0
=> r₁ = 2, r₂ = 4
=> a_n = b₁ * 2ⁿ + b₂ * 4ⁿ
      with the initial condition: a₀ = 4, a₁ = 10
=> 4 = b₁ * 2⁰ + b₂ * 4⁰ = b₁ + b₂
      10 = b₁ * 2¹ + b₂ * 4¹ = 2b₁ + 4b₂
=> b₁ = 3, b₂ = 1
=> a_n = 3 * 2ⁿ + 4ⁿ
 

e) a_n = a_n-2 for n>=2, a₀ = 5, a₁ = -1
Answer: a_n = 2 + 3 * (-1)ⁿ
      c₁ = 0, c₂ = 1,
      r² - 1 = 0
=> r₁ = 1, r₂ = -1
=> a_n = b₁ * 1ⁿ + b₂ * (-1)ⁿ = b₁ + b₂ * (-1)ⁿ
      with the initial condition: a₀ = 5, a₁ = -1
=> 5 = b₁ + b₂ * (-1)⁰ = b₁ + b₂
      -1 = b₁ + b₂ * (-1)¹ = b₁ - b₂
=> b₁ = 2, b₂ = 3
=> a_n = 2 + 3 * (-1)ⁿ
 

 
7. In how many ways can a 2*n rectangular board be tiled using 1*2 and 2*2 pieces?
Answer: Let a_n be the number of ways that 2*n rectangular board can be tiled using 1*2 and 2*2 pieces.
In order to tile the 2*n board from the 2*(n-1) board:

• If the last piece is a 2*2 piece, then we need to find all the ways of tiling 2*(n-2) board, which is a_n-2. Because we can put the 2*2 piece right behind any pattern of 2*(n-2) board.
• If the last piece is a 1*2 piece:
• If the second last one is a 2*2 piece, then we need to find  all the ways of tiling 2*(n-1) board, which is a_n-1. Because we can put the 1*2 piece right behind the 2*2 piece.
• If the second last one is a 1*2 piece, then we need to find all the ways of tiling 2*(n-2) board, which is a_n-2. Because we can rearrange the last two 1*2 pieces by rotating 90 degrees. 

13. Find the solution to a_n = 7a_n-2 + 6a_n-3 with a₀ = 9, a₁ = 10, and a₂ = 32
Answer: a_n = 8 * (-1)ⁿ + 4 * 3ⁿ - 3 * (-2)ⁿ
      c₁ = 0, c₂ = 7, c₃ = 6
      r³ - 7r - 6 = 0
=> r³ - r - 6r - 6 = r(r² - 1) - 6(r + 1) = r(r+1)(r-1) - 6(r+1) = (r+1)(r² - r - 6) = (r+1)(r-3)(r+2) = 0
=> r₁ = -1, r₂ = 3, r₂ = -2
=> a_n = b₁ * (-1)ⁿ + b₂ * 3ⁿ + b₃ * (-2)ⁿ
     with the initial condition: a₀ = 9, a₁ = 10, a₂ = 32
=> 9 = b₁ * (-1)⁰ + b₂ * 3⁰ + b₃ * (-2)⁰  = b₁ + b₂ + b₃
      10 = b₁ * (-1)¹ + b₂ * 3¹ + b₃ * (-2)¹  = -b₁ + 3b₂ - 2b₃
      32 = b₁ * (-1)² + b₂ * 3² + b₃ * (-2)²  = b₁ + 9b₂ + 4b₃
=> b₁ = 8, b₂ = 4, b₂ = -3
=> a_n = 8 * (-1)ⁿ + 4 * 3ⁿ - 3 * (-2)ⁿ
 

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