Further Maths Recurrence relations Practice Questions

Free Further Maths Recurrence relations practice questions with full step-by-step worked solutions. Covers recurrence-relations, iteration, first-order, closed-form. Practise exam-style problems and check your method.

recurrence-relationsiterationfirst-orderclosed-formsecond-orderauxiliary-equation
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A sequence is defined by un+1=3un+2u_{n+1}=3u_{n}+2, n≥1n\ge 1, with u1=4u_{1}=4. Find the value of u4u_{4}.
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Worked solution

  1. Apply the recurrence with n=1n=1

    u2=3⋅4+(2)=14u_{2}=3\cdot 4+\left(2\right)=14

    Each new term is built from the term immediately before it.

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

    u4=3⋅44+(2)=134u_{4}=3\cdot 44+\left(2\right)=134

    Each new term is built from the term immediately before it.

  3. State the required term

    u4=134u_{4}=134

    This is the value asked for.

Answer
134134
Question 2
2 markseasy
A sequence is defined by un+1=12unu_{n+1}=\frac{1}{2}u_{n}, n≥0n\ge 0, with u0=8u_{0}=8. Which of the following best describes the behaviour of unu_{n} for large nn?
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Worked solution

  1. Find the roots that control the behaviour

    m=12m=\frac{1}{2}

    The long-run behaviour is decided entirely by the roots of the auxiliary equation.

  2. Find the modulus of each root

    ∣12∣=12\left|\frac{1}{2}\right|=\frac{1}{2}

    A modulus greater than 11 means growth, less than 11 means decay and exactly 11 means neither.

  3. Generate the first few terms

    u0=8,u1=4,u2=2,u3=1,u4=12,u5=14u_{0}=8,\quad u_{1}=4,\quad u_{2}=2,\quad u_{3}=1,\quad u_{4}=\frac{1}{2},\quad u_{5}=\frac{1}{4}

    The numerical behaviour must agree with what the roots predict.

  4. Select the correct description

    ∣m∣=12 ⇒ un→0\left|m\right|=\frac{1}{2}\ \Rightarrow\ u_{n}\to0

    This is the only option consistent with the roots and with the terms generated above: The terms converge to 0

Answer
The terms converge to 00.
Question 3
4 marksintermediate
A sequence satisfies un+2=6un+1−9unu_{n+2}=6u_{n+1}-9u_{n}. Which of the following is the general solution of this recurrence relation?
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Worked solution

  1. Try a solution of the form un=mnu_{n}=m^{n}

    mn+2=6mn+1−9mnm^{n+2}=6m^{n+1}-9m^{n}

    Substituting un=mnu_{n}=m^{n} into the homogeneous recurrence turns it into an equation for mm.

  2. Divide through by mnm^{n} to obtain the auxiliary equation

    m2−6m+9=0m^{2} - 6 m + 9=0

    Dividing by mnm^{n} (which is never zero) leaves a quadratic in mm.

  3. Solve the auxiliary equation

    (m−3)2=0 ⇒ m=3 (repeated)\left(m - 3\right)^{2}=0\ \Rightarrow\ m=3\ \text{(repeated)}

    The discriminant is zero, so the auxiliary equation has one repeated root.

  4. Build the complementary function from the roots

    un=A3n+Bn⋅3nu_{n}=A3^{n}+Bn\cdot 3^{n}

    This is the general solution: two arbitrary constants for a second-order recurrence.

  5. Reject the option that uses the coefficients as the roots

    6 and −9 are coefficients, not roots6\ \text{and}\ -9\ \text{are coefficients, not roots}

    The roots come from solving the auxiliary equation, not from reading off the recurrence.

  6. Reject the repeated-root form

    (A+Bn)αn is only used when Δ=0\left(A+Bn\right)\alpha^{n}\ \text{is only used when}\ \Delta=0

    Here the discriminant is 00.

  7. Select the general solution

    un=(A+Bn)3nu_{n}=\left(A+Bn\right)3^{n}

    This is the only option that satisfies the recurrence for all values of the arbitrary constants.

Answer
(A+Bn)3n\left(A+Bn\right)3^{n}
Question 4
6 markshard
A sequence satisfies un+2=2un+1−4unu_{n+2}=2u_{n+1}-4u_{n}. Which of the following is the general solution of this recurrence relation?
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Worked solution

  1. Try a solution of the form un=mnu_{n}=m^{n}

    mn+2=2mn+1−4mnm^{n+2}=2m^{n+1}-4m^{n}

    Substituting un=mnu_{n}=m^{n} into the homogeneous recurrence turns it into an equation for mm.

  2. Divide through by mnm^{n} to obtain the auxiliary equation

    m2−2m+4=0m^{2} - 2 m + 4=0

    Dividing by mnm^{n} (which is never zero) leaves a quadratic in mm.

  3. Solve the auxiliary equation

    m=1+3i,m=1−3im=1 + \sqrt{3} i,\qquad m=1 - \sqrt{3} i

    The discriminant is negative, so the roots are a complex conjugate pair.

  4. Write the roots in modulus-argument form

    r=∣m∣=2,θ=arg⁡m=π3r=\left|m\right|=2,\qquad\theta=\arg m=\frac{\pi}{3}

    The complex roots are r(cos⁡θ±isin⁡θ)r\left(\cos\theta\pm i\sin\theta\right), and these are the numbers that appear in the real solution.

  5. Build the complementary function from the roots

    un=2n(Acos⁡(πn3)+Bsin⁡(πn3))u_{n}=2^{n}\left(A\cos\left(\frac{\pi n}{3}\right)+B\sin\left(\frac{\pi n}{3}\right)\right)

    This is the general solution: two arbitrary constants for a second-order recurrence.

  6. Reject the option that uses the coefficients as the roots

    2 and −4 are coefficients, not roots2\ \text{and}\ -4\ \text{are coefficients, not roots}

    The roots come from solving the auxiliary equation, not from reading off the recurrence.

  7. Reject the repeated-root form

    (A+Bn)αn is only used when Δ=0\left(A+Bn\right)\alpha^{n}\ \text{is only used when}\ \Delta=0

    Here the discriminant is −12-12.

  8. Check the option satisfies the recurrence

    un+2−2un+1−−4un=0u_{n+2}-2u_{n+1}--4u_{n}=0

    The correct general solution satisfies the recurrence for every AA and BB.

  9. Recall the auxiliary 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 the homogeneous recurrence and dividing by mnm^{n} gives this quadratic.

  10. Recall the general 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. Select the general solution

    un=2n(Acos⁡(πn3)+Bsin⁡(πn3))u_{n}=2^{n}\left(A\cos\left(\frac{\pi n}{3}\right)+B\sin\left(\frac{\pi n}{3}\right)\right)

    This is the only option that satisfies the recurrence for all values of the arbitrary constants.

Answer
2n(Acos⁡(πn3)+Bsin⁡(πn3))2^{n}\left(A\cos\left(\frac{\pi n}{3}\right)+B\sin\left(\frac{\pi n}{3}\right)\right)
Question 5
9 markschallenging
Let unu_{n} be the number of ways of climbing a flight of nn stairs, taking either 11 stair or 22 stairs at each move. It can be shown that the sequence is defined by un+2=un+1+unu_{n+2}=u_{n+1}+u_{n}, n≥1n\ge 1, with u1=1u_{1}=1 and u2=2u_{2}=2. Find the number of ways of climbing a flight of 1515 stairs.
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Worked solution

  1. Apply the recurrence with n=1n=1

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

    Each new term is built from the terms immediately before it.

  2. Apply the recurrence with n=2n=2

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

    Each new term is built from the terms immediately before it.

  3. Apply the recurrence with n=3n=3

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

    Each new term is built from the terms immediately before it.

  4. Apply the recurrence with n=4n=4

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

    Each new term is built from the terms immediately before it.

  5. Apply the recurrence with n=5n=5

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

    Each new term is built from the terms immediately before it.

  6. Apply the recurrence with n=6n=6

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

    Each new term is built from the terms immediately before it.

  7. Apply the recurrence with n=7n=7

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

    Each new term is built from the terms immediately before it.

  8. Apply the recurrence with n=8n=8

    u10=55+34=89u_{10}=55+34=89

    Each new term is built from the terms immediately before it.

  9. Apply the recurrence with n=9n=9

    u11=89+55=144u_{11}=89+55=144

    Each new term is built from the terms immediately before it.

  10. Apply the recurrence with n=10n=10

    u12=144+89=233u_{12}=144+89=233

    Each new term is built from the terms immediately before it.

  11. Apply the recurrence with n=11n=11

    u13=233+144=377u_{13}=233+144=377

    Each new term is built from the terms immediately before it.

  12. Apply the recurrence with n=12n=12

    u14=377+233=610u_{14}=377+233=610

    Each new term is built from the terms immediately before it.

  13. Apply the recurrence with n=13n=13

    u15=610+377=987u_{15}=610+377=987

    Each new term is built from the terms immediately before it.

  14. List the terms found so far

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

    Writing the terms in a row makes a slip easy to spot.

  15. Note that no closed form is needed

    u15 follows from 13 applications of the recurrenceu_{15}\ \text{follows from 13 applications of the recurrence}

    For a small index, iterating is quicker than solving the recurrence.

  16. State the required term

    u15=987u_{15}=987

    This is the value asked for.

Answer
987987

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