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A first-order linear recurrence relation of the form
\(u_0=a, \quad \quad u_{n+1}=Ru_n+d\)
generates the terms of a sequence. A geometric sequence will be generated if
\(D\)
\(\text{Consider Option D where }R=2\ \text{and }d=0\)
\(u_0=a, \quad u_{n+1}=2u_n\)
\(\text{Generates the geometric sequence }\ a,\ 2a,\ 4a,\ 8a,\ …\)
\(\Rightarrow D\)
A sequence of numbers is generated by the recurrence relation shown below.
\(T_0=5, \quad T_{n+1}=-T_n\)
The value of \(T_2\) is
\(D\)
\(T_1 = -T_0 = -5\)
\(T_2 = -T_1 = 5\)
\(\Rightarrow D\)
`A_n` is the `n`th term in a sequence.
Which one of the following expressions does not define a geometric sequence?
| A. `A_(n + 1) = n` | `\ \ \ \ A_0 = 1` |
| B. `A_(n + 1) = 4` | `\ \ \ \ A_0 = 4` |
| C. `A_(n + 1) = A_n + A_n` | `\ \ \ \ A_0 = 3` |
| D. `A_(n + 1) = –A_n` | `\ \ \ \ A_0 = 5` |
| E. `A_(n + 1) = 4A_n` | `\ \ \ \ A_0 = 2` |
`A`
`text(Test all options by looking at the first)`
`text(3 terms that each produces.)`
`text(Consider)\ A,`
`A_0=1, \ A_1=0, \ A_2=1`
`text(There is no common ratio in this sequence.)`
`text(All other options can be shown to have a common ratio.)`
`=> A`
A town has a population of 200 people when a company opens a large mine.
Due to the opening of the mine, the town’s population is expected to increase by 50% each year.
Let `P_n` be the population of the town `n` years after the mine opened.
The expected growth in the town’s population can be modelled by
| A. `P_(n + 1) = P_n + 100` | `\ \ \ \ \ P_0 = 200` |
| B. `P_(n + 1) = P_n + 100` | `\ \ \ \ \ P_1= 300` |
| C. `P_(n + 1) = 0.5P_n` | `\ \ \ \ \ P_0 = 200` |
| D. `P_(n + 1) = 1.5P_n` | `\ \ \ \ \ P_0 = 300` |
| E. `P_(n + 1) = 1.5P_n` | `\ \ \ \ \ P_1 = 300` |
`E`
`text(After 1 year,)`
| `P_1` | `= 1.5 xx P_0` |
| `= 1.5 xx 200` | |
| `=300` |
`=> E`
The following information relates to Parts 1 and 2.
The number of waterfowl living in a wetlands area has decreased by 4% each year since 2003.
At the start of 2003 the number of waterfowl was 680.
Part 1
If this percentage decrease continues at the same rate, the number of waterfowl in the wetlands area at the start of 2008 will be closest to
A. 532
B. 544
C. 554
D. 571
E. 578
Part 2
`W_n` is the number of waterfowl at the start of the `n`th year.
Let `W_1 = 680.`
The rule for a difference equation that can be used to model the number of waterfowl in the wetlands area over time is
A. `W_(n+1) = W_n - 0.04n`
B. `W_(n+1) = 1.04 W_n`
C. `W_(n+1) = 0.04 W_n`
D. `W_(n+1) = -0.04 W_n`
E. `W_(n+1) = 0.96 W_n`
`text (Part 1:)\ C`
`text (Part 2:)\ E`
`text (Part 1)`
`text(After 1 year, number of waterfowls)`
`=680 – 4/100 xx 680`
`=680\ (0.96)^1`
`text(After 2 years)\ = 680\ (0.96)^2`
`vdots`
`text{After 5 years (in 2008)}`
`=680\ (0.96)^5 =554.45…`
`rArr C`
`text (Part 2)`
`text(Sequence is geometric where)\ \ r=0.96`
| `:. W_(n+1)/W_n` | `=0.96` |
| `W_(n+1)` | `=0.96 W_n` |
`rArr E`
In 2008, there are 800 bats living in a park.
After 2008, the number of bats living in the park is expected to increase by 15% per year.
Let `Β_n` represent the number of bats living in the park `n` years after 2008.
A difference equation that can be used to determine the number of bats living in the park `n` years after 2008 is
| A. `B_n=1.15B_(n-1)-800` | `\ \ \ \ \ B_0=2008` |
| B. `B_n=B_(n-1)+1.15xx800` | `\ \ \ \ \ B_0=2008` |
| C. `B_n=B_(n-1)-0.15xx800` | `\ \ \ \ \ B_0=800` |
| D. `B_n=0.15B_(n-1)` | `\ \ \ \ \ B_0=800` |
| E. `B_n=1.15B_(n-1)` | `\ \ \ \ \ B_0=800` |
`E`
`B_0=800`
`B_1= B_0 + 15 text(%) xx B_0=1.15 B_0`
`B_2= 1.15B_1`
`vdots`
`B_n=1.15 B_(n-1)`
`=> E`
Consider the following sequence.
`2,\ 1,\ 0.5\ …`
Which of the following difference equations could generate this sequence?
| A. | `t_(n + 1) = t_n - 1` | `t_1 = 2` |
| B. | `t_(n + 1) = 3 - t_n` | `t_1 = 2` |
| C. | `t_(n + 1) = 2 × 0.5^(n – 1)` | `t_1 = 2` |
| D. | `t_(n + 1) = - 0.5t_n + 2` | `t_1 = 2` |
| E. | `t_(n + 1) = 0.5t_n` | `t_1 = 2` |
`E`
`text(Sequence is)\ \ 2, 1, 0.5, …`
`=>\ text(Geometric sequence where common ratio = 0.5)`
`∴\ text(Difference equation is)`
`t_(n + 1) = 0.5t_n`
`=> E`
A poultry farmer aims to increase the weight of a turkey by 10% each month.
The turkey’s weight, `T_n`, in kilograms, after `n` months, would be modelled by the rule
A. `T_(n + 1) = T_n + 10`
B. `T_(n + 1) = 1.1T_n + 10`
C. `T_(n + 1) = 0.10T_n`
D. `T_(n + 1) = 10T_n`
E. `T_(n + 1) = 1.1T_n`
`E`
| `T_2` | `=1.1T_1` |
| `T_3` | `= 1.1T_2` |
| `vdots` | |
| `T_(n+1)` | `= 1.1T_n` |
`rArr E`
A sequence is generated by the difference equation
`t_(n+1)=2 xx t_n,\ \ \ \ \ t_1=1`
The `n`th term of this sequence is
A. `t_n=1×2^(n-1)`
B. `t_n=1+2^(n-1)`
C. `t_n=1+2×(n-1)`
D. `t_n=2+(n-1)`
E. `t_n=2+1^(n-1)`
`A`
| `t_2` | `=2 xx t_1 = 2` |
| `t_3` | `=2 xx t_2 = 2^2` |
| `t_4` | `=2 xx t_3 = 2^3` |
| `t_5` | `=2 xx t_4 = 2^4` |
`vdots`
`t_n= 1 xx 2^(n-1)`
`=> A`