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Recursion, GEN1 2024 NHT 18 MC

A sequence of numbers is generated by a recurrence relation of the form

\(T_0=5, \quad T_{n+1}=k-T_n\)

All terms of the sequence have the same value.

The constant \(k\) is equal to

  1. \(-\)10
  2. \(-\)5
  3. 0
  4. 5
  5. 10
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\(E\)

Show Worked Solution

\(\text{By trial and error with each option:}\)

\(\text{If \(k\) = 10,}\)

\(T_1=10-5=5, \ T_2=10-T_1=5, …\)

\(\Rightarrow E\)

Recursion, GEN1 2022 VCAA 17 MC

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

  1. \(-4\)
  2. \(-2\)
  3. \(0\)
  4. \(2\)
  5. \(4\)
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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%.

CORE, FUR1 2021 VCAA 17 MC

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

  1. – 6
  2. – 4
  3.    4
  4.    6
  5.  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`

CORE*, FUR1 2006 VCAA 5 MC

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`

CORE*, FUR1 2009 VCAA 6 MC

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`

Show Worked Solution

`text(By elimination,)`

♦♦ Mean mark 33%.
MARKERS’ COMMENT: Many students clearly did not reason their way through the solution by generating a few terms. This is an efficient and effective strategy.
`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`

CORE*, FUR1 2014 VCAA 4 MC

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`