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Recurrence relations questions

A sequence defined by its own previous terms can be solved outright. The same auxiliary equation that cracked differential equations works here, one step at a time instead of one instant at a time.

6 original questions · 32 marks · the recurrence relations notes · Further Pure 2

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

• 4 marks · warm-up

  Solve u_n+1 = 4u_n − 3 with u₁ = 2.

  Worked answer

  Complementary function: A4ⁿ. For the constant, try u_n = c, so c = 4c − 3 and c = 1. Then u_n = A4ⁿ + 1, and u₁ = 2 gives 4A + 1 = 2, so A = 1/4. Hence u_n = 4ⁿ⁻¹ + 1. B1 for the complementary function, M1 for a constant particular solution, A1 for c = 1, A1 for the final formula. The first four terms are 2, 5, 17, 65 either way.

• 5 marks · warm-up

  Solve u_n+1 = 2u_n + n with u₁ = 1.

  Worked answer

  The right-hand side is linear in n, so try a particular solution an + b, so a(n + 1) + b = 2(an + b) + n. Comparing coefficients of n gives a = 2a + 1, so a = −1; the constants give a + b = 2b, so b = −1. With the complementary function A2ⁿ, u₁ = 1 gives 2A − 2 = 1, so A = 3/2 and u_n = 3(2ⁿ⁻¹) − n − 1. M1 for trying an + b, A1 for a = −1, A1 for b = −1, M1 for using the initial condition, A1 for the final formula. Checking n = 3: 12 − 4 = 8, and the recurrence gives 8 too.

• 6 marks

  Solve u_n+2 = 5u_n+1 − 6u_n with u₁ = 1 and u₂ = 4.

  Worked answer

  Auxiliary equation m² − 5m + 6 = 0, so m = 2 or m = 3 and u_n = A2ⁿ + B3ⁿ. The conditions give 2A + 3B = 1 and 4A + 9B = 4, so B = 2/3 and A = −1/2. Tidying, u_n = 2(3ⁿ⁻¹) − 2ⁿ⁻¹. M1 for the auxiliary equation, A1 for the two roots, B1 for the general solution, M1 for using both conditions, A1 for A and B, A1 for the final formula. Checking n = 3: 18 − 4 = 14, and 5(4) − 6(1) = 14.

• 6 marks

  Solve u_n+2 = 6u_n+1 − 9u_n with u₁ = 3 and u₂ = 18.

  Worked answer

  Auxiliary equation m² − 6m + 9 = 0 has the repeated root m = 3, so the general solution needs the extra factor of n and reads u_n = (A + Bn)3ⁿ. The conditions give 3(A + B) = 3 and 9(A + 2B) = 18, so A + B = 1 and A + 2B = 2, giving B = 1 and A = 0. Hence u_n = n3ⁿ. M1 for the auxiliary equation, A1 for the repeated root, B1 for the general solution with its factor of n, M1 for the two conditions, A1 for A and B, A1 for the final formula. Checking n = 3: 81, and 6(18) − 9(3) = 81.

• 4 marks · prove

  Prove by induction that u_n = 4ⁿ⁻¹ + 1 satisfies u_n+1 = 4u_n − 3 with u₁ = 2.

  Worked answer

  Base case: 4⁰ + 1 = 2, as given. Assume u_k = 4^k−1 + 1. Then u_k+1 = 4(4^k−1 + 1) − 3 = 4^k + 4 − 3 = 4^k + 1, which is the formula at n = k + 1. B1 for the base case, M1 for substituting the hypothesis into the recurrence, A1 for 4^k + 1, A1 for the conclusion. True at n = 1 and inherited at each step, so true for all positive integers n.

• 7 marks · stretch

  Solve u_n+2 + u_n+1 − 6u_n = 12 with u₁ = −4 and u₂ = 10.

  Worked answer

  Auxiliary equation m² + m − 6 = 0 gives m = 2 or m = −3. For the constant 12, try u_n = k, giving k + k − 6k = −4k = 12, so k = −3. Then u_n = A2ⁿ + B(−3)ⁿ − 3, and the conditions give 2A − 3B = −1 and 4A + 9B = 13, so A = 1 and B = 1. Hence u_n = 2ⁿ + (−3)ⁿ − 3. M1 for the auxiliary equation, A1 for the two roots, M1 for a constant particular solution, A1 for k = −3, M1 for the two conditions, A1 for A and B, A1 for the final formula. Checking n = 3: 8 − 27 − 3 = −22, and 12 − 10 + 6(−4) = −22.

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