Hard Further Maths Recurrence relations Questions

Challenging, exam-style Further Maths Recurrence relations questions with worked solutions. Stretch yourself on the hardest recurrence-relations, first-order, modelling, savings problems.

recurrence-relationsfirst-ordermodellingsavingsloansecond-order
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
A drug-infusion model each hour keeps 30%30\% of the current dose and a steady 1414 milligrams is infused; unu_{n} is the amount, in milligrams, after nn hours. The model is un+1=310un+14u_{n+1}=\frac{3}{10}u_{n}+14 for n≥0n\ge 0, with u0=5u_{0}=5. Find the long-run amount LL that unu_{n} approaches.
Show worked solution

Worked solution

  1. Set both terms equal to the steady value LL

    L=3L10+14L = \frac{3 L}{10} + 14

    If the model settles then consecutive terms tend to the same value, so the limit is a fixed point of the recurrence.

  2. Solve for the steady state

    7L10=14 ⇒ L=20\frac{7 L}{10}=14\ \Rightarrow\ L=20

    Collecting the LL terms gives one linear equation.

  3. Justify that the model really settles

    ∣310∣<1\left|\frac{3}{10}\right|<1

    The transient part A(310)nA\left(\frac{3}{10}\right)^{n} decays because the growth factor has modulus below 11.

  4. Step through a few values as a check

    u0=5, u1=312, u2=37320, u3=3919200u_{0}=5,\ u_{1}=\frac{31}{2},\ u_{2}=\frac{373}{20},\ u_{3}=\frac{3919}{200}

    The values are visibly closing in on the fixed point.

  5. Note the closed-form structure

    un=A(310)n+20u_{n}=A\left(\frac{3}{10}\right)^{n}+20

    The geometric part dies away and only the fixed point survives.

  6. Recall how a recurrence generates a sequence

    un+1 (or un+2) is built from the earlier term(s)u_{n+1}\ \text{(or }u_{n+2}\text{) is built from the earlier term(s)}

    A recurrence relation says how to get the next term from the ones before it, so the whole sequence follows from the starting value(s).

  7. Recall the auxiliary (characteristic) equation

    un+2=aun+1+bun ⇒ m2−am−b=0u_{n+2}=au_{n+1}+bu_{n}\ \Rightarrow\ m^{2}-am-b=0

    Trying un=mnu_{n}=m^{n} in a homogeneous second-order recurrence and dividing by mnm^{n} gives this quadratic.

  8. Recall the solution for two distinct real roots

    un=Aαn+Bβnu_{n}=A\alpha^{n}+B\beta^{n}

    Each root of the auxiliary equation contributes one geometric term.

  9. Recall the solution for a repeated root

    un=(A+Bn)αnu_{n}=\left(A+Bn\right)\alpha^{n}

    A repeated root supplies one solution, so the second carries a factor nn.

  10. Recall the solution for complex roots

    un=rn(Acos⁡nθ+Bsin⁡nθ)u_{n}=r^{n}\left(A\cos n\theta+B\sin n\theta\right)

    Writing the conjugate roots as r(cos⁡θ±isin⁡θ)r\left(\cos\theta\pm i\sin\theta\right) turns the solution into a real oscillation.

  11. Recall the fixed point of a first-order model

    L=aL+b ⇒ L=b1−a(a≠1)L=aL+b\ \Rightarrow\ L=\frac{b}{1-a}\quad\left(a\neq1\right)

    A model that settles down must settle to a fixed point of its recurrence.

  12. Recall the condition for a model to settle

    ∣a∣<1 for the geometric part to die away\left|a\right|<1\ \text{for the geometric part to die away}

    A growth factor of modulus 11 or more makes the transient persist or explode.

  13. Recall the first-order homogeneous solution

    un+1=aun ⇒ un=Aanu_{n+1}=au_{n}\ \Rightarrow\ u_{n}=Aa^{n}

    A constant multiplier each step produces a geometric sequence.

  14. Recall how many starting values are needed

    order 1→1 value,order 2→2 values\text{order }1\rightarrow 1\ \text{value},\qquad\text{order }2\rightarrow 2\ \text{values}

    Exactly that many initial conditions are needed to pin the model down.

  15. Recall the standard trial particular solutions

    f(n)=c⇒λ; f(n)=cn+d⇒λn+μ; f(n)=ckn⇒λknf(n)=c\Rightarrow\lambda;\ f(n)=cn+d\Rightarrow\lambda n+\mu;\ f(n)=ck^{n}\Rightarrow\lambda k^{n}

    The particular solution copies the shape of the forcing term f(n)f(n).

  16. State the long-run value

    L=20L=20

    This is the value the model approaches as nn grows.

Answer
2020
Question 2
9 markschallenging
A depreciating fleet keeps 90%90\% of its value and a fixed £1200\pounds1200 maintenance credit is added each year; unu_{n} is the net value, in pounds, after nn years. The model is un+1=910un+1200u_{n+1}=\frac{9}{10}u_{n}+1200 for n≥0n\ge 0, with u0=8000u_{0}=8000. Find a formula for unu_{n} in terms of nn.
Show worked solution

Worked solution

  1. Solve the homogeneous part of the model

    un+1=910un ⇒ un=A(910)nu_{n+1}=\frac{9}{10}u_{n}\ \Rightarrow\ u_{n}=A\left(\frac{9}{10}\right)^{n}

    Ignoring the constant term leaves a geometric sequence with ratio 910\frac{9}{10}.

  2. Find the fixed point for the particular part

    L=910L+1200 ⇒ L=12000L=\frac{9}{10}L+1200\ \Rightarrow\ L=12000

    A constant forcing term is matched by the constant fixed point of the model.

  3. Fit the constant to the starting value

    u0=8000 ⇒ un=−4000(910)n+12000u_{0}=8000\ \Rightarrow\ u_{n}=-4000\left(\frac{9}{10}\right)^{n}+12000

    The one starting value determines the single arbitrary constant.

  4. Check the formula on the first few terms

    u0=8000, u1=8400, u2=8760, u3=9084u_{0}=8000,\ u_{1}=8400,\ u_{2}=8760,\ u_{3}=9084

    The closed form must reproduce the values the model generates.

  5. Note the model being solved

    un+1=910un+1200u_{n+1}=\frac{9}{10}u_{n}+1200

    The closed form has to satisfy this relation for every nn.

  6. Recall how a recurrence generates a sequence

    un+1 (or un+2) is built from the earlier term(s)u_{n+1}\ \text{(or }u_{n+2}\text{) is built from the earlier term(s)}

    A recurrence relation says how to get the next term from the ones before it, so the whole sequence follows from the starting value(s).

  7. Recall the auxiliary (characteristic) equation

    un+2=aun+1+bun ⇒ m2−am−b=0u_{n+2}=au_{n+1}+bu_{n}\ \Rightarrow\ m^{2}-am-b=0

    Trying un=mnu_{n}=m^{n} in a homogeneous second-order recurrence and dividing by mnm^{n} gives this quadratic.

  8. Recall the solution for two distinct real roots

    un=Aαn+Bβnu_{n}=A\alpha^{n}+B\beta^{n}

    Each root of the auxiliary equation contributes one geometric term.

  9. Recall the solution for a repeated root

    un=(A+Bn)αnu_{n}=\left(A+Bn\right)\alpha^{n}

    A repeated root supplies one solution, so the second carries a factor nn.

  10. Recall the solution for complex roots

    un=rn(Acos⁡nθ+Bsin⁡nθ)u_{n}=r^{n}\left(A\cos n\theta+B\sin n\theta\right)

    Writing the conjugate roots as r(cos⁡θ±isin⁡θ)r\left(\cos\theta\pm i\sin\theta\right) turns the solution into a real oscillation.

  11. Recall the fixed point of a first-order model

    L=aL+b ⇒ L=b1−a(a≠1)L=aL+b\ \Rightarrow\ L=\frac{b}{1-a}\quad\left(a\neq1\right)

    A model that settles down must settle to a fixed point of its recurrence.

  12. Recall the condition for a model to settle

    ∣a∣<1 for the geometric part to die away\left|a\right|<1\ \text{for the geometric part to die away}

    A growth factor of modulus 11 or more makes the transient persist or explode.

  13. Recall the first-order homogeneous solution

    un+1=aun ⇒ un=Aanu_{n+1}=au_{n}\ \Rightarrow\ u_{n}=Aa^{n}

    A constant multiplier each step produces a geometric sequence.

  14. Recall how many starting values are needed

    order 1→1 value,order 2→2 values\text{order }1\rightarrow 1\ \text{value},\qquad\text{order }2\rightarrow 2\ \text{values}

    Exactly that many initial conditions are needed to pin the model down.

  15. State the closed form

    un=−4000(910)n+12000u_{n}=-4000\left(\frac{9}{10}\right)^{n}+12000

    This formula gives any term of the model directly from nn.

Answer
−4000(910)n+12000-4000\left(\frac{9}{10}\right)^{n}+12000
Question 3
9 markschallenging
A two-tap echo filter unu_{n} satisfies a second-order model with distinct real growth factors. Here unu_{n} satisfies un+2=un+1+6unu_{n+2}=u_{n+1}+6u_{n}. Which of the following is the general solution of this recurrence relation?
Show worked solution

Worked solution

  1. Form the auxiliary equation

    m2−m−6=0m^{2} - m - 6=0

    Trying un=mnu_{n}=m^{n} turns the homogeneous model into this quadratic.

  2. Solve for the roots

    m=−2, m=3m=-2,\ m=3

    The roots decide the shape of the general solution.

  3. Match the roots to the correct solution shape

    un=A⋅3n+B(−2)nu_{n}=A\cdot 3^{n}+B\left(-2\right)^{n}

    Two distinct real roots give two independent geometric terms.

  4. Reject options using the coefficients as roots

    coefficients are not the roots of the auxiliary equation\text{coefficients are not the roots of the auxiliary equation}

    The roots come from solving the quadratic, not from reading off the model.

  5. Check the chosen solution satisfies the model

    un+2−un+1+6un=0u_{n+2}-u_{n+1}+6u_{n}=0

    The correct general solution satisfies the model for all values of the constants.

  6. Recall how a recurrence generates a sequence

    un+1 (or un+2) is built from the earlier term(s)u_{n+1}\ \text{(or }u_{n+2}\text{) is built from the earlier term(s)}

    A recurrence relation says how to get the next term from the ones before it, so the whole sequence follows from the starting value(s).

  7. Recall the auxiliary (characteristic) equation

    un+2=aun+1+bun ⇒ m2−am−b=0u_{n+2}=au_{n+1}+bu_{n}\ \Rightarrow\ m^{2}-am-b=0

    Trying un=mnu_{n}=m^{n} in a homogeneous second-order recurrence and dividing by mnm^{n} gives this quadratic.

  8. Recall the solution for two distinct real roots

    un=Aαn+Bβnu_{n}=A\alpha^{n}+B\beta^{n}

    Each root of the auxiliary equation contributes one geometric term.

  9. Recall the solution for a repeated root

    un=(A+Bn)αnu_{n}=\left(A+Bn\right)\alpha^{n}

    A repeated root supplies one solution, so the second carries a factor nn.

  10. Recall the solution for complex roots

    un=rn(Acos⁡nθ+Bsin⁡nθ)u_{n}=r^{n}\left(A\cos n\theta+B\sin n\theta\right)

    Writing the conjugate roots as r(cos⁡θ±isin⁡θ)r\left(\cos\theta\pm i\sin\theta\right) turns the solution into a real oscillation.

  11. Recall the fixed point of a first-order model

    L=aL+b ⇒ L=b1−a(a≠1)L=aL+b\ \Rightarrow\ L=\frac{b}{1-a}\quad\left(a\neq1\right)

    A model that settles down must settle to a fixed point of its recurrence.

  12. Recall the condition for a model to settle

    ∣a∣<1 for the geometric part to die away\left|a\right|<1\ \text{for the geometric part to die away}

    A growth factor of modulus 11 or more makes the transient persist or explode.

  13. Recall the first-order homogeneous solution

    un+1=aun ⇒ un=Aanu_{n+1}=au_{n}\ \Rightarrow\ u_{n}=Aa^{n}

    A constant multiplier each step produces a geometric sequence.

  14. Recall how many starting values are needed

    order 1→1 value,order 2→2 values\text{order }1\rightarrow 1\ \text{value},\qquad\text{order }2\rightarrow 2\ \text{values}

    Exactly that many initial conditions are needed to pin the model down.

  15. Recall the standard trial particular solutions

    f(n)=c⇒λ; f(n)=cn+d⇒λn+μ; f(n)=ckn⇒λknf(n)=c\Rightarrow\lambda;\ f(n)=cn+d\Rightarrow\lambda n+\mu;\ f(n)=ck^{n}\Rightarrow\lambda k^{n}

    The particular solution copies the shape of the forcing term f(n)f(n).

  16. Recall the resonance rule

    if the trial already solves the homogeneous part, multiply it by n\text{if the trial already solves the homogeneous part, multiply it by }n

    Otherwise the substitution collapses to 0=f(n)0=f(n) and no coefficient works.

  17. Select the general solution

    un=A⋅3n+B(−2)nu_{n}=A\cdot 3^{n}+B\left(-2\right)^{n}

    This option satisfies the model for all values of the arbitrary constants.

Answer
A⋅3n+B(−2)nA\cdot 3^{n}+B\left(-2\right)^{n}
Question 4
9 markschallenging
A compound-and-drain reservoir each week retains 150%150\% of its level and then 2020 units drain away; unu_{n} is the level after nn weeks and satisfies un+1=32un−20u_{n+1}=\frac{3}{2}u_{n}-20. The model is un+1=32un−20u_{n+1}=\frac{3}{2}u_{n}-20 for n≥0n\ge 0, with u0=60u_{0}=60. Find the level after 55 weeks, that is the value of u5u_{5}.
Show worked solution

Worked solution

  1. Apply the model with n=0n=0

    u1=32⋅60+(−20)=70u_{1}=\frac{3}{2}\cdot 60+\left(-20\right)=70

    Each new value comes straight from the model applied to the value just before it.

  2. Apply the model with n=1n=1

    u2=32⋅70+(−20)=85u_{2}=\frac{3}{2}\cdot 70+\left(-20\right)=85

    Each new value comes straight from the model applied to the value just before it.

  3. Apply the model with n=2n=2

    u3=32⋅85+(−20)=2152u_{3}=\frac{3}{2}\cdot 85+\left(-20\right)=\frac{215}{2}

    Each new value comes straight from the model applied to the value just before it.

  4. Apply the model with n=3n=3

    u4=322152+(−20)=5654u_{4}=\frac{3}{2}\frac{215}{2}+\left(-20\right)=\frac{565}{4}

    Each new value comes straight from the model applied to the value just before it.

  5. Apply the model with n=4n=4

    u5=325654+(−20)=15358u_{5}=\frac{3}{2}\frac{565}{4}+\left(-20\right)=\frac{1535}{8}

    Each new value comes straight from the model applied to the value just before it.

  6. List the values found so far

    u0=60, u1=70, u2=85, u3=2152, u4=5654u_{0}=60,\ u_{1}=70,\ u_{2}=85,\ u_{3}=\frac{215}{2},\ u_{4}=\frac{565}{4}

    Writing the run of values in order makes an arithmetic slip easy to spot.

  7. Note the model itself

    un+1=32un−20u_{n+1}=\frac{3}{2}u_{n}-20

    Every step is one substitution into this relation.

  8. Recall how a recurrence generates a sequence

    un+1 (or un+2) is built from the earlier term(s)u_{n+1}\ \text{(or }u_{n+2}\text{) is built from the earlier term(s)}

    A recurrence relation says how to get the next term from the ones before it, so the whole sequence follows from the starting value(s).

  9. Recall the auxiliary (characteristic) equation

    un+2=aun+1+bun ⇒ m2−am−b=0u_{n+2}=au_{n+1}+bu_{n}\ \Rightarrow\ m^{2}-am-b=0

    Trying un=mnu_{n}=m^{n} in a homogeneous second-order recurrence and dividing by mnm^{n} gives this quadratic.

  10. Recall the solution for two distinct real roots

    un=Aαn+Bβnu_{n}=A\alpha^{n}+B\beta^{n}

    Each root of the auxiliary equation contributes one geometric term.

  11. Recall the solution for a repeated root

    un=(A+Bn)αnu_{n}=\left(A+Bn\right)\alpha^{n}

    A repeated root supplies one solution, so the second carries a factor nn.

  12. Recall the solution for complex roots

    un=rn(Acos⁡nθ+Bsin⁡nθ)u_{n}=r^{n}\left(A\cos n\theta+B\sin n\theta\right)

    Writing the conjugate roots as r(cos⁡θ±isin⁡θ)r\left(\cos\theta\pm i\sin\theta\right) turns the solution into a real oscillation.

  13. Recall the fixed point of a first-order model

    L=aL+b ⇒ L=b1−a(a≠1)L=aL+b\ \Rightarrow\ L=\frac{b}{1-a}\quad\left(a\neq1\right)

    A model that settles down must settle to a fixed point of its recurrence.

  14. Recall the condition for a model to settle

    ∣a∣<1 for the geometric part to die away\left|a\right|<1\ \text{for the geometric part to die away}

    A growth factor of modulus 11 or more makes the transient persist or explode.

  15. Recall the first-order homogeneous solution

    un+1=aun ⇒ un=Aanu_{n+1}=au_{n}\ \Rightarrow\ u_{n}=Aa^{n}

    A constant multiplier each step produces a geometric sequence.

  16. State the required value

    u5=15358u_{5}=\frac{1535}{8}

    This is the quantity the model is asked for.

Answer
15358\frac{1535}{8}
Question 5
9 markschallenging
A honeybee ancestry counts, in unu_{n}, the ancestors nn generations back of a male bee; the counts satisfy a Fibonacci recurrence. The model is un+2=un+1+unu_{n+2}=u_{n+1}+u_{n} for n≥1n\ge 1, with u1=1u_{1}=1 and u2=2u_{2}=2. Find the number of ancestors 1515 generations back, that is the value of u15u_{15}.
Show worked solution

Worked solution

  1. Apply the model with n=1n=1

    u3=2+1=3u_{3}=2+1=3

    Each new value comes straight from the model applied to the values just before it.

  2. Apply the model with n=2n=2

    u4=3+2=5u_{4}=3+2=5

    Each new value comes straight from the model applied to the values just before it.

  3. Apply the model with n=3n=3

    u5=5+3=8u_{5}=5+3=8

    Each new value comes straight from the model applied to the values just before it.

  4. Apply the model with n=4n=4

    u6=8+5=13u_{6}=8+5=13

    Each new value comes straight from the model applied to the values just before it.

  5. Apply the model with n=5n=5

    u7=13+8=21u_{7}=13+8=21

    Each new value comes straight from the model applied to the values just before it.

  6. Apply the model with n=6n=6

    u8=21+13=34u_{8}=21+13=34

    Each new value comes straight from the model applied to the values just before it.

  7. Apply the model with n=7n=7

    u9=34+21=55u_{9}=34+21=55

    Each new value comes straight from the model applied to the values just before it.

  8. List the values found so far

    u1=1, u2=2, u3=3, u4=5, u5=8u_{1}=1,\ u_{2}=2,\ u_{3}=3,\ u_{4}=5,\ u_{5}=8

    Writing the run of values in order makes an arithmetic slip easy to spot.

  9. Note the model itself

    un+2=un+1+unu_{n+2}=u_{n+1}+u_{n}

    Every step is one substitution into this relation.

  10. Recall how a recurrence generates a sequence

    un+1 (or un+2) is built from the earlier term(s)u_{n+1}\ \text{(or }u_{n+2}\text{) is built from the earlier term(s)}

    A recurrence relation says how to get the next term from the ones before it, so the whole sequence follows from the starting value(s).

  11. Recall the auxiliary (characteristic) equation

    un+2=aun+1+bun ⇒ m2−am−b=0u_{n+2}=au_{n+1}+bu_{n}\ \Rightarrow\ m^{2}-am-b=0

    Trying un=mnu_{n}=m^{n} in a homogeneous second-order recurrence and dividing by mnm^{n} gives this quadratic.

  12. Recall the solution for two distinct real roots

    un=Aαn+Bβnu_{n}=A\alpha^{n}+B\beta^{n}

    Each root of the auxiliary equation contributes one geometric term.

  13. Recall the solution for a repeated root

    un=(A+Bn)αnu_{n}=\left(A+Bn\right)\alpha^{n}

    A repeated root supplies one solution, so the second carries a factor nn.

  14. Recall the solution for complex roots

    un=rn(Acos⁡nθ+Bsin⁡nθ)u_{n}=r^{n}\left(A\cos n\theta+B\sin n\theta\right)

    Writing the conjugate roots as r(cos⁡θ±isin⁡θ)r\left(\cos\theta\pm i\sin\theta\right) turns the solution into a real oscillation.

  15. State the required value

    u15=987u_{15}=987

    This is the quantity the model is asked for.

Answer
987987

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