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Sequences and recurrence relationsAQA A-Level Maths: Flashcards

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What is a recurrence relation?

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What is a recurrence relation?
A rule xn+1=f(xn)x_{n+1}=f(x_n) giving each term from the previous one, with a starting value.
u1=3u_1=3, un+1=2un−1u_{n+1}=2u_n-1: first five terms?
3,5,9,17,333,5,9,17,33
When is a sequence increasing?
un+1>unu_{n+1}>u_n for all nn.
When is a sequence decreasing?
un+1<unu_{n+1}<u_n for all nn.
How do you prove a sequence is increasing?
Show un+1−un>0u_{n+1}-u_n>0 for all nn.
What is a periodic sequence?
One where un+k=unu_{n+k}=u_n for all nn; the same kk terms repeat.
What is the order of a periodic sequence?
The number of terms in one cycle.
x1=2x_1=2, xn+1=11−xnx_{n+1}=\frac{1}{1-x_n}: its terms and order?
2,−1,12,2,…2,-1,\frac12,2,\dots, order 33
How do you find x100x_{100} in a sequence of order 33?
100=3×33+1100=3\times33+1, so x100=x1x_{100}=x_1.
How do you sum many terms of a periodic sequence?
Sum one cycle, multiply by the number of whole cycles, add the leftover terms.
Check that un=2n+1u_n=2^n+1 fits un+1=2un−1u_{n+1}=2u_n-1.
2(2n+1)−1=2n+1+1=un+12(2^n+1)-1=2^{n+1}+1=u_{n+1}
an=nn+2a_n=\frac{n}{n+2}: is it increasing?
Yes, an+1−an=2(n+2)(n+3)>0a_{n+1}-a_n=\frac{2}{(n+2)(n+3)}>0.

Exam questions on Sequences and recurrence relations

  1. A sequence is defined by u1=3u_1=3 and un+1=2un−1u_{n+1}=2u_n-1 for n≥1n\ge1.
    Show that un=2n+1u_n=2^n+1 satisfies both u1=3u_1=3 and the relation un+1=2un−1u_{n+1}=2u_n-1.2 marks
  2. A sequence has nnth term an=nn+2a_n=\frac{n}{n+2} for n≥1n\ge1.
    Prove that the sequence is increasing.2 marks
  3. A sequence is defined by x1=2x_1=2 and xn+1=11−xnx_{n+1}=\frac{1}{1-x_n} for n≥1n\ge1.
    Find x2x_2, x3x_3 and x4x_4, and state what this shows about the sequence.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).