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Recurrence relations questions

A sequence defined in terms of its own earlier terms can be solved for an explicit formula. The auxiliary equation used for second order differential equations works here too, with the roots governing the general solution and initial terms fixing the constants.

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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