Model with recursive sequences

Learning goals

Generates sequence terms from a recursive rule and initial conditions.
Interprets each part of the recurrence relation.
Builds a recurrence relation from a repeated process or context.
Compares recursive and explicit descriptions when possible.

Difficulty: 1 – 5

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Model with recursive sequences 1

A plant is $6\,\text{cm}$ tall at the end of week $1$. In a simple model, its height increases by $4\,\text{cm}$ each week. The model is $h_1=6$ and $h_n=h_{n-1}+4$ for $n\ge2$. Find $h_2,h_3,h_4,h_5$, and explain what the $+4$ means in the recurrence.

Hints

- Start with the given initial value before using the recursive rule. - Each new value depends on the immediately preceding value. - Interpret the added constant in the units and time interval of the model.

Solution

Apply the recurrence one week at a time: $h_2=10$, $h_3=14$, $h_4=18$, and $h_5=22$. The $+4$ means the model adds $4\,\text{cm}$ to the previous week's height to obtain the next week's height.

Answer

$h_2=10\,\text{cm}$, $h_3=14\,\text{cm}$, $h_4=18\,\text{cm}$, $h_5=22\,\text{cm}$. The $+4$ represents weekly growth of $4\,\text{cm}$.

Model with recursive sequences 2

A video has $80$ views at the end of hour $1$. In a simple model, the total number of views is multiplied by $1.5$ each hour. The model is $V_1=80$, $V_n=1.5V_{n-1}$ for $n\ge2$. Find $V_2$ and $V_3$, and explain what the factor $1.5$ means in this recurrence.

Hints

- Start with the given initial value and apply the recursive multiplier one step at a time. - Interpret a factor greater than $1$ as multiplicative growth from one step to the next.

Solution

1. Apply the recurrence once: $V_2=1.5\cdot80=120$. 2. Apply it again: $V_3=1.5\cdot120=180$. 3. The factor $1.5$ means each modeled total is $150\%$ of the previous total, which is a $50\%$ increase each hour.

Answer

$V_2=120$, $V_3=180$. The factor $1.5$ means each hour's modeled total is $1.5$ times the previous total, a $50\%$ increase.

Model with recursive sequences 3

Ridgeview Robotics Club has $12$ members in week $1$. The club gains exactly $5$ members each week in a simple model. Write a recursive definition for the modeled number of members $R_n$. Include the initial value and the rule for later weeks. Do not calculate later terms.

Hints

- A recursive definition needs a starting value before it can generate later values. - Translate “gains $5$ members each week” into a rule relating one week to the previous week.

Solution

1. The initial value is $R_1=12$. 2. Each later week has $5$ more members than the preceding week, so $R_n=R_{n-1}+5$ for $n\ge2$.

Answer

$R_1=12$, $R_n=R_{n-1}+5$ for $n\ge2$.

Model with recursive sequences 4

An art class starts with $x_0=10\,\text{yd}$ of ribbon. Each day, the class uses half the ribbon it has and then receives $8\,\text{yd}$ more. Find $x_1$ and write a recursive rule for $x_{n+1}$ using $x_n$.

Hints

- Which amount must you update: the original length or the length left at the start of that day? - In which order should the two daily changes be applied? - How could one expression represent the same process on any day?

Solution

1. Half the starting ribbon remains, then the delivery is added: $x_1=0.5\cdot10+8=13\,\text{yd}$. 2. The same steps apply each day, so $x_{n+1}=0.5x_n+8$ for $n\ge0$, with $x_0=10$, where $x_n$ is measured in yards.

Answer

$x_1=13\,\text{yd}$; $x_{n+1}=0.5x_n+8$ for $n\ge0$, with $x_0=10$, in yards.

Model with recursive sequences 5

Each day, a greenhouse water tank loses $25\%$ of its water and then receives $10$ gallons. Let $W_n$ be the amount after the addition on day $n$. Which rule fits the order of events? Explain. A: $W_{n+1}=0.75W_n+10$ B: $W_{n+1}=0.75(W_n+10)$

Hints

- Compare what each rule does to the newly added water. - Test both rules on an empty tank. Which result follows the stated order of events?

Solution

1. After losing $25\%$, the tank keeps $75\%$ of its previous amount: $0.75W_n$. 2. The $10$ gallons are added afterward, so rule A, $W_{n+1}=0.75W_n+10$, is correct. Rule B would remove water from the new addition too.

Answer

A: $W_{n+1}=0.75W_n+10$; the addition happens after the loss.

Model with recursive sequences 6

An account starts with $\$1000$ and earns $5\%$ interest at the end of each year, with no deposits or withdrawals. Let $B_n$ be the balance after $n$ years. a) Write a recursive model with an initial value. b) Write an equivalent explicit formula. c) Find $B_3$ to the nearest cent.

Hints

- Convert the percent increase into the multiplier applied to the previous balance. - In the recursive form, distinguish the starting balance from the update rule. - In the explicit form, the exponent counts how many annual multiplications have occurred.

Solution

a) The initial balance is $B_0=1000$. Each year the balance is multiplied by $1.05$, so $B_n=1.05B_{n-1}$ for $n\ge1$. b) Repeated multiplication by $1.05$ gives $B_n=1000(1.05)^n$. c) $B_3=1000(1.05)^3=1157.625$, so the balance is $\$1157.63$.

Answer

a) $B_0=1000$, $B_n=1.05B_{n-1}$ for $n\ge1$ b) $B_n=1000(1.05)^n$ c) $\$1157.63$

Model with recursive sequences 7

A calculator app repeats the operation in the diagram. It adds the same fixed amount each time. Let $x_0$ be the first input. Find the fixed amount added and the number that replaces the question mark. Write the recursive update as $x_{n+1}$ in terms of $x_n$.

Hints

- What relationship must connect the first input and its output? - Once you know the fixed addition, what should count as the input to the second step? - How can you describe the repeated operation without using the particular numbers in the diagram?

Solution

1. Halving $40$ gives $20$. To reach $23$, the fixed amount added must be $3$. 2. Repeat with $23$: $0.5\cdot23+3=14.5$. 3. The repeated update is $x_{n+1}=0.5x_n+3$, starting at $x_0=40$.

Answer

Fixed amount added: $3$; missing value: $14.5$; $x_0=40$, $x_{n+1}=0.5x_n+3$ for $n\ge0$.

Model with recursive sequences 8

A school fundraiser starts with $\$30$. Each week, it spends $20\%$ of the balance, then receives a $\$10$ donation. Paul says the balance after both week $1$ and week $2$ is $\$34$. Find the correct balances for both weeks and explain where his work went wrong.

Hints

- Identify the balance at the end of the first week. - What amount does the fundraiser have when the second week begins? - Check Paul's second-week result against the stated weekly process.

Solution

1. After week $1$, the balance is $0.8\cdot30+10=34$ dollars. 2. Week $2$ must start with the new balance: $0.8\cdot34+10=37.20$ dollars. 3. Paul reused the original amount instead of using the result from the previous week.

Answer

Week $1$: $\$34$; week $2$: $\$37.20$. Paul should have started week $2$ with $\$34$, not $\$30$.

Model with recursive sequences 9

A cat shelter has $40$ pounds of dry food. Each day it uses $25\%$ of the food on hand and then gets a delivery of $c$ pounds. After the first day, it has $38$ pounds. Find $c$ and write a recursive rule for the amount of food $F_n$ after day $n$, starting with $F_0=40$.

Hints

- What fraction of the food is left after one day's use? - Which part of the first day's change is not yet known? - How can the same daily steps be written using the current amount of food?

Solution

1. After the shelter uses $25\%$, it has $0.75\cdot40=30$ pounds left. 2. The delivery adds $c=38-30=8$ pounds. 3. The repeated rule is $F_{n+1}=0.75F_n+8$, with $F_0=40$ and $n\ge0$.

Answer

$c=8\,\text{lb}$; $F_0=40$, $F_{n+1}=0.75F_n+8$ for $n\ge0$, with $F_n$ in pounds.

Model with recursive sequences 10

An acting club and a music club track their warm-up times before rehearsal. Each club either adds the same number of minutes each week or multiplies its time by the same factor each week. <table> <tr><th>Week</th><th>Acting club</th><th>Music club</th></tr> <tr><td>$0$</td><td>$6$</td><td>$6$</td></tr> <tr><td>$1$</td><td>$10$</td><td>$9$</td></tr> <tr><td>$2$</td><td>$14$</td><td>$13.5$</td></tr> </table> Write a recursive rule for each club's warm-up time. Explain how you can tell which plan adds the same number of minutes and which uses the same multiplier.

Hints

- Compare how each club's practice time changes from one week to the next. - Check whether subtraction or division gives the same result at each step. - Use week $0$ as the starting value of each rule.

Solution

1. The acting club's planned times increase by $4$ minutes each week: $10-6=14-10=4$. Write $A_0=6$, $A_{n+1}=A_n+4$ for $n\ge0$. 2. The music club's planned times are each $1.5$ times the previous time: $9\div6=13.5\div9=1.5$. Write $M_0=6$, $M_{n+1}=1.5M_n$ for $n\ge0$. 3. The acting club uses a fixed increase in minutes; the music club uses a fixed factor.

Answer

Acting club: $A_0=6$, $A_{n+1}=A_n+4$ for $n\ge0$; differences are $10-6=14-10=4$ minutes. Music club: $M_0=6$, $M_{n+1}=1.5M_n$ for $n\ge0$; ratios are $\frac{9}{6}=\frac{13.5}{9}=1.5$.

Model with recursive sequences 11

A water tank holds up to $160$ gallons. In each filling cycle, the tank takes in the same fraction of the space still empty. a) What fraction of the previous cycle's empty space remains after each new cycle? b) How many gallons of water will be in the tank after cycle $4$? c) Use the pattern in the empty space to explain why the model stays below $160$ gallons after every finite number of cycles.

Hints

- Compare the heights of two neighboring points on the graph. - Continue the pattern for one more cycle before finding the water amount. - How does the empty space relate to the tank's capacity and the water already in it?

Solution

1. The graph gives $64,32,16,8$ gallons of empty space. Each is half the previous amount, so the fraction remaining is $\frac{1}{2}$. 2. After cycle $4$, the tank has room for another $8\cdot\frac{1}{2}=4$ gallons. It contains $160-4=156$ gallons. 3. The remaining space is positive at every finite cycle, so the predicted water amount stays below $160$ gallons.

Answer

a) $\frac{1}{2}$ b) $156\,\text{gal}$ c) The empty space starts positive and is halved each cycle, so it never reaches zero in a finite number of cycles. The amount of water stays below $160\,\text{gal}$.

Model with recursive sequences 12

A greenhouse model predicts $8\,\text{°C}$ on day $1$ and $20\,\text{°C}$ on day $2$. Each later prediction is the average of the previous two days' temperatures. a) Find the predicted temperatures on days $3$, $4$, and $5$. b) Write a recursive rule for the temperature $T_n$ on day $n$, for $n\ge3$. c) Explain why the day-$1$ temperature alone is not enough to predict day $3$.

Hints

- Work forward one term at a time and use the most recent two values. - Write an average of two neighboring terms using their subscripts. - Ask what information is missing if you know only the first term.

Solution

1. The day-$3$ prediction is $T_3=\frac{8+20}{2}=14\,\text{°C}$. 2. Day $4$ gives $T_4=\frac{20+14}{2}=17\,\text{°C}$, and day $5$ gives $T_5=\frac{14+17}{2}=15.5\,\text{°C}$. 3. The rule is $T_n=\frac{T_{n-1}+T_{n-2}}{2}$ for $n\ge3$, with $T_1=8$ and $T_2=20$. 4. Two starting temperatures are needed because each new prediction depends on the two days before it.

Answer

a) Day $3$: $14\,\text{°C}$; day $4$: $17\,\text{°C}$; day $5$: $15.5\,\text{°C}$ b) $T_1=8$, $T_2=20$, $T_n=\frac{T_{n-1}+T_{n-2}}{2}$ for $n\ge3$ c) Two starting temperatures are needed because the rule uses the two previous days.

Model with recursive sequences 13

An indoor fountain starts with $30\,\text{gal}$ of water. Each day it loses $25\%$ of the water it has, then gets $6\,\text{gal}$ more. a) Write a recursive rule for the water amount $W_n$ after day $n$, with $W_0=30$. b) Find the amount $L$ that would stay the same after one full day. c) Find $W_1$. Is it closer to $L$ than the starting amount is?

Hints

- What fraction of the water remains before the extra water arrives? - For an amount to stay unchanged, what must be true about the amount before and after a day? - Compare each amount's distance from the unchanged level, not just the amounts themselves.

Solution

1. Each day, $75\%$ of the current water remains before $6\,\text{gal}$ is added. Thus $W_{n+1}=0.75W_n+6$, with $W_0=30$ and $n\ge0$. 2. An unchanged amount satisfies $L=0.75L+6$, so $0.25L=6$ and $L=24\,\text{gal}$. 3. After one day, $W_1=0.75\cdot30+6=28.5\,\text{gal}$. The starting amount is $6\,\text{gal}$ above $L$, while $W_1$ is only $4.5\,\text{gal}$ above it.

Answer

a) $W_0=30$, $W_{n+1}=0.75W_n+6$ for $n\ge0$, in gallons. b) $L=24\,\text{gal}$ c) $W_1=28.5\,\text{gal}$; yes, it is closer to $L$.

Model with recursive sequences 14

A bottle of warm water starts at $50\,\text{°C}$ in a room at $18\,\text{°C}$. In a simple cooling model, the difference between the water and room temperatures is cut in half every hour. a) Write a recursive rule for the water temperature $T_n$ after $n$ hours, starting with $T_0=50$. b) Find $T_1$ and $T_2$. c) Explain why this model keeps the water above $18\,\text{°C}$ and makes it cooler each hour.

Hints

- How can you describe the water's temperature by its distance above room temperature? - Why should the recursive rule include the room's temperature? - How does multiplying a positive temperature difference by the same positive fraction affect its sign?

Solution

1. After an hour, half the difference above room temperature remains. Thus $T_{n+1}=18+0.5(T_n-18)=0.5T_n+9$, with $T_0=50$ and $n\ge0$. 2. The first two predictions are $T_1=18+0.5(50-18)=34\,\text{°C}$ and $T_2=18+0.5(34-18)=26\,\text{°C}$. 3. A positive difference above $18\,\text{°C}$ stays positive when halved. It also becomes smaller each hour, so the model stays above room temperature while cooling toward it.

Answer

a) $T_0=50$, $T_{n+1}=18+0.5(T_n-18)$ for $n\ge0$, in degrees Celsius. b) $T_1=34\,\text{°C}$; $T_2=26\,\text{°C}$ c) Its positive difference above $18\,\text{°C}$ halves each hour, so it stays positive but gets smaller.

Model with recursive sequences 15

A patient takes a $20\,\text{mg}$ tablet once each day at the same time. During each day, the body eliminates $30\%$ of the medication currently present. a) Make a table of the amount in the body immediately after each dose for days $1$ through $5$. Before the first dose, the amount is $0\,\text{mg}$. b) On what day does the amount first exceed $50\,\text{mg}$? c) Find the theoretical long-term amount in the body immediately after a dose.

Hints

- At each step, keep the stated fraction of the previous amount before adding the new dose. - Compare the generated values with the threshold only after the recurrence is applied. - At equilibrium, one full recurrence step leaves the amount unchanged.

Solution

a) If $x_n$ is the amount immediately after the dose on day $n$, then $x_n=0.7x_{n-1}+20$, with $x_0=0$. Thus, $x_1=20$, $x_2=34$, $x_3=43.8$, $x_4=50.66$, and $x_5=55.462$, all in milligrams. b) The first value greater than $50\,\text{mg}$ is $x_4=50.66\,\text{mg}$, so the amount first exceeds the threshold on day $4$. c) At equilibrium, $L=0.7L+20$. Thus $0.3L=20$, so $L=\frac{200}{3}\approx66.67\,\text{mg}$.

Answer

a) <table> <tr> <th>Day</th> <th>Amount after dose (mg)</th> </tr> <tr><td>$1$</td><td>$20$</td></tr> <tr><td>$2$</td><td>$34$</td></tr> <tr><td>$3$</td><td>$43.8$</td></tr> <tr><td>$4$</td><td>$50.66$</td></tr> <tr><td>$5$</td><td>$55.462$</td></tr> </table> b) Day $4$ c) $\frac{200}{3}\,\text{mg}\approx66.67\,\text{mg}$

Model with recursive sequences 16

Immediately after the final dose of an earlier treatment, a patient has $40\,\text{mg}$ of a medication in the body at day $0$. The patient now takes $5\,\text{mg}$ once daily. Before each new dose, the body eliminates $20\%$ of the medication present. a) Find the amount immediately after the dose on days $1$ and $2$. b) Show that the amount decreases over time despite the daily dose, and find the value to which it stabilizes. c) How would the limiting amount change if the body eliminated $25\%$ each day instead of $20\%$?

Hints

- Compare the amount removed during a day with the fixed amount added afterward. - To prove the direction of change, express $A_{n+1}-A_n$ in terms of $A_n$. - Compare each term with the equilibrium value to determine whether the sequence can cross it. - In part c, change only the retention factor in the equilibrium equation.

Solution

1. The recurrence is $A_{n+1}=0.8A_n+5$, with $A_0=40$. Thus $A_1=0.8\cdot40+5=37\,\text{mg}$ and $A_2=0.8\cdot37+5=34.6\,\text{mg}$. 2. The daily change is $A_{n+1}-A_n=5-0.2A_n$, which is negative whenever $A_n>25$. 3. Also, $A_{n+1}-25=0.8(A_n-25)$. Since $A_0>25$, every term remains above $25$, so the change stays negative while the distance from $25$ is multiplied by $0.8$. Therefore, the sequence decreases toward $25\,\text{mg}$. 4. With a $25\%$ daily elimination, the limit satisfies $L=0.75L+5$, giving $L=20\,\text{mg}$. The limiting amount decreases by $5\,\text{mg}$.

Answer

a) Day $1$: $37\,\text{mg}$; day $2$: $34.6\,\text{mg}$ b) $A_{n+1}-A_n=5-0.2A_n<0$ while $A_n>25$, and $A_{n+1}-25=0.8(A_n-25)>0$. Thus the amount decreases while staying above $25\,\text{mg}$, and it approaches $25\,\text{mg}$. c) The limiting amount would decrease to $20\,\text{mg}$.

Model with recursive sequences 17

A newly planted forest can support at most $4000$ trees. Initially, $800$ young trees are planted. Each year, the number of trees increases by $25\%$ of the remaining capacity, the difference between $4000$ and the current number of trees. a) Write a recurrence for $B_{n+1}$ in terms of $B_n$. b) Find the model values for the first five years after planting. Round to the nearest whole tree. c) Explain why the model represents bounded growth and describe how the annual increases behave over time.

Hints

- Express the remaining capacity as the difference between the maximum and the current amount. - Add the stated fraction of that remaining capacity to the current value. - For the long-run behavior, focus on what happens to the gap from capacity.

Solution

a) The annual increase is $0.25(4000-B_n)$, so $B_{n+1}=B_n+0.25(4000-B_n)=0.75B_n+1000$, with $B_0=800$. b) Without rounding intermediate values, $B_1=1600$, $B_2=2200$, $B_3=2650$, $B_4=2987.5\approx2988$, and $B_5=3240.625\approx3241$. c) The population is bounded above by $4000$. As the gap to capacity becomes smaller, the annual increase $0.25(4000-B_n)$ also becomes smaller and approaches $0$.

Answer

a) $B_{n+1}=B_n+0.25(4000-B_n)$, or $B_{n+1}=0.75B_n+1000$ b) $B_1=1600$, $B_2=2200$, $B_3=2650$, $B_4\approx2988$, $B_5\approx3241$ c) The capacity $4000$ is an upper bound, and the annual increases approach $0$.

Model with recursive sequences 18

Two exercise plans predict workout times, in minutes: Plan I: $x_0=10$, $x_n=x_{n-1}+4$ for $n\ge1$ Plan II: $y_0=10$, $y_n=1.5y_{n-1}$ for $n\ge1$ a) Match each graph to Plan I or Plan II. b) Explain how the weekly changes in workout time support your match.

Hints

- Compare the vertical gaps between neighboring dots in each graph. - If a growing workout time is multiplied by the same factor each week, what happens to the increases between weeks? - What feature of a graph would rule out a plan that adds the same number of minutes each week?

Solution

1. Graph a) matches Plan I: the times are $10,14,18,22,26,30$ minutes and increase by $4$ minutes each week. 2. Graph b) matches Plan II: the times are $10,15,22.5,33.75,50.625,75.9375$ minutes and each is $1.5$ times the previous time. 3. In graph a), the vertical gaps stay equal. In graph b), they increase because adding $50\%$ of a growing amount gives a larger increase each week.

Answer

a) Graph a): Plan I; graph b): Plan II. b) Plan I adds $4$ minutes each week. Plan II increases by $50\%$ each week, so its weekly increases grow.

Model with recursive sequences 19

In a simplified wildlife model, a reserve begins with $2$ nesting pairs in month $1$ and $3$ nesting pairs in month $2$. For every later month, the predicted number of nesting pairs equals the sum of the predictions from the previous two months. a) Write a recursive definition for $P_n$. b) Find $P_3$ through $P_7$. c) Find $P_8$, and explain why this model cannot be written as either $P_n=P_{n-1}+d$ with one constant $d$ or $P_n=rP_{n-1}$ with one constant $r$.

Hints

- A recurrence that depends on two earlier values needs two initial values. - Generate each new month only after the two values immediately before it are known. - To compare this model with one-step additive or multiplicative rules, inspect consecutive differences and consecutive ratios.

Solution

1. The recursive definition is $P_1=2$, $P_2=3$, and $P_n=P_{n-1}+P_{n-2}$ for $n\ge3$. 2. Applying the recurrence gives $P_3=5$, $P_4=8$, $P_5=13$, $P_6=21$, and $P_7=34$. 3. Then $P_8=P_7+P_6=34+21=55$. 4. The consecutive differences are not constant, so no single additive constant $d$ generates the model. The consecutive ratios are also not constant, so no single multiplicative constant $r$ generates it. The next value depends on two previous values rather than only one.

Answer

a) $P_1=2$, $P_2=3$, $P_n=P_{n-1}+P_{n-2}$ for $n\ge3$ b) $P_3=5$, $P_4=8$, $P_5=13$, $P_6=21$, $P_7=34$ c) $P_8=55$; the differences and ratios are not constant, and each new term depends on the previous two terms.

Model with recursive sequences 20

An artificial pond begins empty. Each day, $15\%$ of the water evaporates, and then $60\,\text{L}$ of fresh water is added. a) Find the amount of water immediately after the addition on days $2$, $3$, and $4$. b) Find the theoretical upper limit of the water amount. c) On what day does the amount first exceed $250\,\text{L}$?

Hints

- Write one recurrence step using the retained water and the fixed addition. - At the limiting amount, one full day leaves the amount unchanged. - For the threshold, compare consecutive day values near the target.

Solution

a) Let $W_n$ be the amount immediately after the addition on day $n$. Then $W_{n+1}=0.85W_n+60$, with $W_0=0$. This gives $W_1=60$, $W_2=111$, $W_3=154.35$, and $W_4=191.1975\approx191.20$, in liters. b) At equilibrium, $L=0.85L+60$. Therefore, $0.15L=60$, so $L=400\,\text{L}$. c) The explicit form is $W_n=400(1-0.85^n)$. It gives $W_6\approx249.14\,\text{L}$ and $W_7\approx271.77\,\text{L}$, so the amount first exceeds $250\,\text{L}$ on day $7$.

Answer

a) Day $2$: $111\,\text{L}$; day $3$: $154.35\,\text{L}$; day $4$: about $191.20\,\text{L}$ b) $400\,\text{L}$ c) Day $7$

Model with recursive sequences 21

Each month, a car-sharing service expects to sign up $20\%$ of the town's residents who are not users yet. The dashed line $s$ marks the town's population. The plotted user counts are model estimates before rounding, not counts of actual people. a) Read the town's population from the graph. Then find the number of users after each of the first three months and complete the table. <table> <tr><th>Month $x$</th><td>$0$</td><td>$1$</td><td>$2$</td><td>$3$</td></tr> <tr><th>Users $N(x)$</th><td>$2000$</td><td>$\square$</td><td>$\square$</td><td>$\square$</td></tr> </table> b) Write an explicit function for the number of users after $x$ months. c) About how many users does the model predict after one year? Round to the nearest person. How do the plotted estimates compare with the population line? Use your formula to explain why the model stays below that line in every finite month.

Hints

- Does each month's $20\%$ apply to the same number of residents? Check the first two updates. - An explicit rule gives the number of users in month $x$ without calculating every earlier month. What part of the monthly update might be easier to describe this way? - What would have to become zero for the model to reach the town's population line?

Solution

1. The dashed line is at $20{,}000$, so the town has $20{,}000$ residents. 2. Each month, add $20\%$ of the remaining nonusers. Month $1$: $2000+0.20\cdot(20{,}000-2000)=5600$. Month $2$: $5600+0.20\cdot(20{,}000-5600)=8480$. Month $3$: $8480+0.20\cdot(20{,}000-8480)=10{,}784$. 3. The number of nonusers begins at $18{,}000$ and is multiplied by $0.8$ each month. Therefore, $N(x)=20{,}000-18{,}000(0.8)^x$ for whole-number months $x\ge0$. 4. After one year, $N(12)=20{,}000-18{,}000(0.8)^{12}\approx18{,}763$ users, rounded to a whole person. The later plotted estimates stay below the dashed line and move closer to it. 5. Because $18{,}000(0.8)^x>0$ for every finite whole-number $x$, the unrounded model has $N(x)<20{,}000$. It never reaches the population at a finite month.

Answer

a) Population: $20{,}000$; month $1$: $5600$; month $2$: $8480$; month $3$: $10{,}784$ b) $N(x)=20{,}000-18{,}000(0.8)^x$ for whole-number $x\ge0$ c) About $18{,}763$ users; later estimates approach the population from below, and $N(x)<20{,}000$ for every finite whole-number month.

Model with recursive sequences 22

A school help desk has one worker. The worker helps one student at a time in arrival order. Arrival times are minutes after the desk opens. The minutes each student needs for help are also shown. <table><tr><th>Student</th><th>Arrival time</th><th>Minutes needed</th></tr><tr><td>1</td><td>$0$</td><td>$3$</td></tr><tr><td>2</td><td>$2$</td><td>$4$</td></tr><tr><td>3</td><td>$3$</td><td>$2$</td></tr><tr><td>4</td><td>$7$</td><td>$5$</td></tr><tr><td>5</td><td>$8$</td><td>$1$</td></tr></table> Let $A_n$ be student $n$'s arrival time, $S_n$ the minutes needed, and $F_n$ the time the worker finishes helping that student. Set $F_0=0$. a) Write a recursive rule for $F_n$ using $A_n$, $S_n$, and $F_{n-1}$. b) Use your rule to find all five finish times and the minutes each student waits before help begins. c) Find the mean waiting time.

Hints

- When is the worker free to begin helping the next student? - What two times determine when each student's help can start? - Once you know the next finish time, how could you work out the time spent waiting? - How do the five waiting times combine into one mean?

Solution

1. A student's help begins at the later of the arrival time and the previous finish time. Thus $F_n=\max(A_n,F_{n-1})+S_n$, with $F_0=0$. 2. Apply the rule in order: $F_1=3$, $F_2=7$, $F_3=9$, $F_4=14$, and $F_5=15$ minutes after opening. 3. Each waiting time is $\max(A_n,F_{n-1})-A_n$. The waits are $0,1,4,2,6$ minutes. 4. Their sum is $13$ minutes, so the mean is $\frac{13}{5}=2.6$ minutes.

Answer

a) $F_0=0$ and $F_n=\max(A_n,F_{n-1})+S_n$. b) Finish times: $3,7,9,14,15$ minutes; waiting times: $0,1,4,2,6$ minutes. c) $2.6$ minutes.

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Model with recursive sequences

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