A sequence of numbers is generated by the recurrence relation shown below.

`P_0 = 2,quadqquadP_(n + 1) = 3P_n - 1`

What is the value of `P_3`?

- 2
- 5
- 11
- 41
- 122

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A sequence of numbers is generated by the recurrence relation shown below.

`P_0 = 2,quadqquadP_(n + 1) = 3P_n - 1`

What is the value of `P_3`?

- 2
- 5
- 11
- 41
- 122

Show Answers Only

`D`

Show Worked Solution

`P_1 = 3P_0 – 1 = 3 xx 2 – 1 = 5`

`P_2 = 3 xx 5 – 1 = 14`

`P_3 = 3 xx 14 – 1 = 41`

`=>\ D`

Consider the recurrence relation shown below.

`A_0 = 3, qquad A_(n + 1) = 2A_n + 4`

The value of `A_3` in the sequence generated by this recurrence relation is given by

- `2 xx 3 + 4`
- `2 xx 4 + 4`
- `2 xx 10 + 4`
- `2 xx 24 + 4`
- `2 xx 52 + 4`

Show Answers Only

`D`

Show Worked Solution

`A_1 = 2A_0 + 4 = 2 xx 3 + 4 = 10`

`A_2 = 2 xx 10 + 4 = 24`

`A_3 = 2 xx 24 + 4`

`=> D`

The first five terms of a sequence are 2, 6, 22, 86, 342 …

The recurrence relation that generates this sequence could be

- `P_0 = 2,qquadP_(n + 1) = P_n + 4`
- `P_0 = 2,qquadP_(n + 1) = 2 P_n + 2`
- `P_0 = 2,qquadP_(n + 1) = 3 P_n`
- `P_0 = 2,qquadP_(n + 1) = 4 P_n – 2`
- `P_0 = 2,qquadP_(n + 1) = 5 P_n – 4`

Show Answers Only

`D`

Show Worked Solution

`text(Using trial and error,)`

`text(Consider option)\ D:`

`P_1 = 4 xx 2 – 2 = 6`

`P_2 = 4 xx 6 – 2 = 22`

`P_3 = 4 xx 22 – 2 = 86\ \ text(etc…)`

`=> D`

Consider the recurrence relation below.

`A_0 = 2,\ \ \ \ \ A_(n + 1) = 3 A_n + 1`

The first four terms of this recurrence relation are

- `0, 2, 7, 22\ …`
- `1, 2, 7, 22\ …`
- `2, 5, 16, 49\ …`
- `2, 7, 18, 54\ …`
- `2, 7, 22, 67\ …`

Show Answers Only

`E`

Show Worked Solution

`A_0` | `= 2\ \ (text(given))` |

`A_1` | `= 3(2) + 1 = 7` |

`A_2` | `= 3(7) + 1 = 22` |

`A_3` | `= 3(22) + 1 = 67` |

`=> E`

Paul has to replace 3000 m of fencing on his farm.

Let `F_n` be the length, in metres, of fencing left to replace after `n` weeks.

The difference equation

`F_(n + 1) = 0.95F_n + a\ \ \ \ \ \ F_0 = 3000`

can be used to calculate the length of fencing left to replace after `n` weeks.

In this equation, `a` is a constant.

After one week, Paul still has 2540 m of fencing left to replace.

After three weeks, the length of fencing, in metres, left to replace will be closest to

**A.** 1310

**B. ** 1380

**C.** 1620

**D.** 1690

**E. ** 2100

Show Answers Only

`D`

Show Worked Solution

`F_(n + 1) = 0.95F_n + a`

`F_1` | `= 0.95F_0 + a` |

`2540` | `= 0.95 xx 3000 + a` |

`:.a` | `= – 310` |

`F_2` | `= 0.95 xx 2540 – 310` |

`= 2103` |

`F_3` | `= 0.95 xx 2103 – 310` |

`= 1687.85` |

`=> D`

Miki is competing as a runner in a half-marathon.

After 30 minutes, his progress in the race is modelled by the difference equation

`K_(n + 1) = 0.99K_n + 250,\ \ \ \ \ \ K_30 = 7550`

where `n ≥ 30` and `K_n` is the total distance Miki has run, in metres, after `n` minutes.

Using this difference equation, the total distance, in metres, that Miki is expected to have run 32 minutes after the start of the race is closest to

**A. ** 7650

**B.** 7725

**C.** 7800

**D.** 7900

**E.** 8050

Show Answers Only

`D`

Show Worked Solution

`K_(n + 1) = 0.99K_n + 250`

`text(S)text(ince)\ \ K_30 = 7550,`

`K_31` | `= 0.99 xx 7550 + 250` |

`= 7724.5` | |

`K_32` | `= 0.99 xx 7724.5 + 250` |

`= 7897.255` |

`=> D`

*The following information relates to Parts 1 and 2.*

A farmer plans to breed sheep to sell.

In the first year she starts with 50 breeding sheep.

During the first year, the sheep numbers increase by 84%.

At the end of the first year, the farmer sells 40 sheep.

**Part 1**

How many sheep does she have at the start of the second year?

**A.** 2

**B.** 42

**C.** 52

**D.** 84

**E.** 92

**Part 2**

If `S_n` is the number of sheep at the start of year `n`*,* a difference equation that can be used to model the growth in sheep numbers over time is

A. `S_(n+1) = 1.84S_n - 40` |
`\ \ \ \ \ text(where)\ \ S_1 = 50` | |

B. `S_(n+1) = 0.84S_n - 50` |
`\ \ \ \ \ text(where)\ \ S_1 = 40` | |

C. `S_(n+1) = 0.84S_n - 40` |
`\ \ \ \ \ text(where)\ \ S_1 = 50` | |

D. `S_(n+1) = 0.16S_n - 50` |
`\ \ \ \ \ text(where)\ \ S_1 = 40` | |

E. `S_(n+1) = 0.16S_n - 40` |
`\ \ \ \ \ text(where)\ \ S_1 = 50` |

Show Answers Only

`text (Part 1:)\ C`

`text (Part 2:)\ A`

Show Worked Solution

`text (Part 1)`

`text(Number at the start of the 2nd year)`

`=50 + (84text(%) xx 50) – 40`

`=52`

`rArr C`

`text (Part 2)`

`text(If sheep numbers increase by 84% from the start of)`

`text(each year, and 40 sheep are then sold,)`

`S_(n+1)` | `=S_n + 0.84S_n – 40` |

`=1.84 S_n – 40` |

`rArr A`

The difference equation

`t_(n+1) = at_n + 6 quad text (where) quad t_1 = 5`

generates the sequence

`5, 21, 69, 213\ …`

The value of `a` is

**A.** – 1

**B.** 3

**C.** 4

**D.** 15

**E.** 16

Show Answers Only

`B`

Show Worked Solution

`t_2` | `=a t_1 +6` |

`:. 21` | `=5a + 6` |

`a` | `=3` |

`rArr B`

The initial rate of pay for a job is $10 per hour.

A worker’s skill increases the longer she works on this job. As a result, the hourly rate of pay increases each month.

The hourly rate of pay in the `n`th month of working on this job is given by the difference equation

`S_(n+1) = 0.2 xx S_n+15\ \ \ \ \ \ S_1 = 10`

The maximum hourly rate of pay that the worker can earn in this job is closest to

**A.** $3.00

**B.** $12.00

**C.** $12.50

**D.** $18.75

**E.** $75.00

Show Answers Only

`D`

Show Worked Solution

`text(Maximum rate)\ S_(max)\ text(occurs when)`

`S_(n+1)` | `=S_n` |

`S_(max)` | `=0.2×S_(max)+15` |

`0.8 xx S_(max)` | `=15` |

`S_(max)` | `=15/0.8` |

`=18.75` |

`=> D`

A patient takes 15 milligrams of a prescribed drug at the start of each day.

Over the next 24 hours, 85% of the drug in his body is used. The remaining 15% stays in his body.

Let `D_n` be the number of milligrams of the drug in the patient’s body immediately after taking the drug at the start of the `n`th day.

A difference equation for determining `D_(n+1)`, the number of milligrams in the patient’s body immediately after taking the drug at the start of the `n+1`th day, is given by

A. `D_(n + 1) = 85 D_n + 15` |
`D_1 = 15` |

B. `D_(n + 1) = 0.85 D_n + 15` |
`D_1 = 15` |

C. `D_(n + 1)= 0.15 D_n + 15` |
`D_1 = 15` |

D. `D_(n + 1)= 0.15 D_n + 0.85` |
`D_1 = 15` |

E. `D_(n + 1)= 15 D_n + 85` |
`D_1 = 15` |

Show Answers Only

`C`

Show Worked Solution

`D_1=15`

`text(85% of the drug is used up before the second dose.)`

`D_2` | `=0.15 D_1 + 15\ \ \ text{(drug left from 1st day + new dose)}` |

`D_3` | `= 0.15 D_2 + 15\ \ \ text{(drug left from 2nd day + new dose)}` |

`vdots` | |

`D_(n+1)` | `=0.15 D_n+15` |

`=> C`

The difference equation `u_(n + 1) = 4u_n - 2` generates a sequence.

If `u_2 = 2`, then `u_4` will be equal to

**A.** 4

**B.** 8

**C.** 22

**D.** 40

**E. ** 42

Show Answers Only

`C`

Show Worked Solution

`u_(n+1)` | `= 4u_n – 2` | |

`∴ u_3` | `= 4u_2 – 2` | |

`= 4 xx 2 – 2\ \ text{(given}\ u_2 = 2 text{)}` | ||

`= 6` | ||

`∴ u_4` | `= 4u_3 – 2` | |

`= 4 xx 6 – 2` | ||

`= 22` |

`=> C`

Let `P_2011` be the number of pairs of shoes that Sienna owns at the end of 2011.

At the beginning of 2012, Sienna plans to throw out the oldest 10% of pairs of shoes that she owned in 2011.

During 2012 she plans to buy 15 new pairs of shoes to add to her collection.

Let `P_2012` be the number of pairs of shoes that Sienna owns at the end of 2012.

A rule that enables `P_2012` to be determined from `P_2011` is

**A.** `P_2012 = 1.1 P_2011 + 15`

**B.** `P_2012 = 1.1 (P_2011 + 15)`

**C.** `P_2012 = 0.1 P_2011 + 15`

**D.** `P_2012 = 0.9 (P_2011 + 15)`

**E.** `P_2012 = 0.9 P_2011 + 15`

Show Answers Only

`E`

Show Worked Solution

`text(By throwing out 10%, Sienna keeps 90% of her)`

`text{her 2011 shoes (or 0.9} \ P_2011 text{) and then adds 15.}`

`:. P_(2012) = 0.9\ P_2011 + 15`

`=> E`

Sam takes a tablet containing 200 mg of medicine once every 24 hours.

Every 24 hours, 40% of the medicine leaves her body. The remaining 60% of the medicine stays in her body.

Let `D_n` be the number of milligrams of the medicine in Sam’s body immediately after she takes the `n`th tablet.

The difference equation that can be used to determine the number of milligrams of the medicine in Sam’s body immediately after she takes each tablet is shown below.

`D_(n + 1) = 0.60D_n + 200,\ \ \ \ \ \ D_1 = 200`

Which one of the following statements is **not** true?

**A.** The number of milligrams of the medicine in Sam’s body never exceeds 500.

**B. **Immediately after taking the third tablet, 392 mg of the medicine is in Sam’s body.

**C.** The number of milligrams of the medicine that leaves Sam’s body during any 24-hour period will always be less than 200.

**D.** The number of milligrams of the medicine that leaves Sam’s body during any 24-hour period is constant.

**E.** If Sam stopped taking the medicine after the fifth tablet, the amount of the medicine in her body would drop to below 200 mg after a further 48 hours.

Show Answers Only

`D`

Show Worked Solution

`text(Consider A:)`

`text(Maximum medicine in body when)`

`D_(n+1)` | `= D_n` |

`x` | `= 0.6x + 200` |

`0.4x` | `= 200` |

`x` | `=500,\ =>\ text(A true)` |

`text(Consider B:)`

`D_1 = 200`

`D_2 = 0.6(200) + 200 = 320`

`D_3 = 0.6(320) + 200 = 392,\ =>\ text(B true)`

`text(Consider C:)`

`text{Max medicine never exceeds 500 mg (from A),}\ =>\ text(C true)`

`text(Consider D:)`

`text(Medicine leaving body is 40% of a changing number,)\ =>\ text(D not true)`

`text(Consider E:)`

`D_4 = 0.6(392) + 200 = 435.2`

`D_5 = 0.6(435.2) + 200 = 461.12`

`D_6 = 0.6(461.12) = 276.67`

`D_7 = 0.6(276.672) = 166.00,\ =>\ text(E true)`

`=>D`

Three years after observations began, 12 300 birds were living in a wetland.

The number of birds living in the wetland changes from year to year according to the difference equation

`t_(n+ 1) = 1.5t_n - 3000, quad quad t_3 = text (12 300)`

where `t_n` is the number of birds observed in the wetland `n` years after observations began.

The number of birds living in the wetland one year after observations began was closest to

**A. ** `8800`

**B. ** `9300`

**C.** `10\ 200`

**D. **`12\ 300`

**E. **`120\ 175`

Show Answers Only

`A`

Show Worked Solution

`t_(n+1)` | `= 1.5t_n – 3000, and t_3 =12\ 300` |

`:. t_3` | `= 1.5t_2 – 3000` |

`12\ 300` | `= 1.5t_2 – 3000` |

`1.5t_2` | `= 15\ 300` |

`t_2` | `= 10\ 200` |

`text(Similarly,)`

`t_2` | `= 1.5t_1 – 3000` |

`10\ 200` | `= 1.5t_1 – 3000` |

`1.5t_1` | `= 13\ 200` |

`:. t_1` | `= 8800` |

`rArr A`