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

\(R_0 = 2,\ \ \ R_{n+1} = 2-R_n\)

The value of \(R_2\) is

- \(-4\)
- \(-2\)
- \(0\)
- \(2\)
- \(4\)

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

\(R_0 = 2,\ \ \ R_{n+1} = 2-R_n\)

The value of \(R_2\) is

- \(-4\)
- \(-2\)
- \(0\)
- \(2\)
- \(4\)

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\(D\)

Show Worked Solution

\(R_0\) | \(=2\ \ \ \ \ R_{n+1} = 2-R_n\) | |

\(R_1\) | \(= 2-2 = 0\) | |

\(R_2\) | \(= 2-0 = 2\) |

\(\Rightarrow D\)

♦ Mean mark 44%.

The following recurrence relation can generate a sequence of numbers.

`L_0 = 37 , \ L_{n+1} = L_n + C`

The value of `L_2` is 25.

The value of `C` is

- – 6
- – 4
- 4
- 6
- 37

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`A`

Show Worked Solution

`L_{n+1} = L_n + C => text{recurrence with with common difference}`

`L_1` | `=37 + C` | |

`L_2` | `=L_1+C` | |

`25` | `=37 + C + C` | |

`2C` | `=-12` | |

`:.C` | `=-6` |

`=> A`

The following recurrence relation can generate a sequence of numbers.

`T_0 = 10, qquad T_(n + 1) = T_n + 3`

The number 13 appears in this sequence as

- `T_1`
- `T_2`
- `T_3`
- `T_10`
- `T_13`

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`A`

Show Worked Solution

`text(If)\ \ n = 0:`

`T_1` | `= T_0 + 3` |

`= 10 + 3` | |

`= 13` |

`=> A`

A difference equation is defined by

`f_(n+1) - f_n = 5\ \ \ \ \ text (where)\ \ f_1 =– 1`

The sequence `f_1, \ f_2, \ f_3, ...` is

**A.** `5, 4, 3\ …`

**B.** `4, 9, 14\ …`

**C.** `– 1, – 6, – 11\ …`

**D.** `– 1, 4, 9\ …`

**E.** `– 1, 6, 11\ …`

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`D`

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`text(Using)\ \ \ f_(n+1) = f_n +5`

`text(Consider)\ \ f_1 = – 1,`

`f_2= – 1 + 5=4`

`f_3 = 4+5=9`

`rArr D`

The `n`th term of a sequence is given by `t_n = 100 − 20n`, where `n = 1, 2, 3, 4\ . . .`

A difference equation that generates the same sequence is

A. `t_(n +1)= 100 - 20t_n` |
`t_1 = 80` |

B. `t_(n+1) = 100t_n - 20` |
`t_1 = 1` |

C. `t_(n+1) = 80t_n` |
`t_1 = 80` |

D. `t_(n+1) = 100 - t_n` |
`t_1 = 20` |

E. `t_(n+1) = t_n - 20` |
`t_1 = 80` |

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`E`

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`text(By elimination,)`

`t_n` | `= 100 − 20_n text( for ) n = 1, 2, …` |

`t_1` | `= 100 \ – 20 × 1` |

`= 80` |

`∴\ text(Eliminate)\ B\ text(and)\ D`

`t_2` | `= 100 − 20 × 2` |

`= 60` |

`∴\ text(Eliminate)\ A\ text(and)\ C`

`=> E`

On day 1, Vikki spends 90 minutes on a training program.

On each following day, she spends 10 minutes less on the training program than she did the day before.

Let `t_n` be the number of minutes that Vikki spends on the training program on day `n`.

A difference equation that can be used to model this situation for `1 ≤ n ≤ 10` is

A. `t_(n + 1) = 0.90t_n` |
`t_1 = 90` |

B. `t_(n + 1) = 1.10 t_n` |
`t_1 = 90` |

C. `t_(n + 1) = t_n - 0.10` |
`t_1 = 90` |

D. `t_(n + 1) = 1 - 10 t_n` |
`t_1 = 90` |

E. `t_(n + 1) = t_n - 10` |
`t_1 = 90` |

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`E`

Show Worked Solution

`text(Difference equation where each term is 10 minutes)`

`text(less than the preceding term.)`

`∴\ text(Equation)\ \ \t_(n+1) = t_n-10, t_1 = 90`

`=> E`