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31 March, 11:23

Consider the sequence defined recursively by an+1 = (an - 1 if an ≥ 10 2an if an < 10)

(a) Let a0 be equal to the last digit in your student number, and compute a1, a2, a3, a4.

(b) Suppose an = 1, and find an+4.

(c) If a0 = 3, does limn→[infinity] an exist?

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  1. 31 March, 11:42
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    a) a₁ = 14, a₂ = 13, a₃ = 12, a₄ = 11

    b) an+4 = 16

    c) Does not exist

    Step-by-step explanation:

    The last digit in student number is given as 7.

    a) a₀ = 7

    Since an < 10 we use 2an

    Therefore a₁ = a₀₊₁ = 2 * a₀ = 2 * 7 = 14

    a₂ = a₁₊₁ = a₁ - 1 = 14 - 1 = 13

    a₃ = a₂₊₁ = a₂ - 1 = 13 - 1 = 12

    a₄ = a₃₊₁ = a₃ - 1 = 12 - 1 = 11

    b) an = 1, we have an+1 = 2an

    Therefore an+2 = an+1+1 = 2 * 2 = 4

    an+3 = an+2+1 = 2 * 4 = 8

    an+4 = an+3+1 = 2 * 8 = 16

    Therefore, an+4 = 16

    c) If a₀ = 3, therefore a₁ = a₀₊₁ = 2*3 = 6

    a₂ = a₁₊₁ = 2*6 = 12

    a₃ = a₂₊₁ = a₂₋₁ = 12 - 1 = 11

    a₄ = a₃₊₁ = a₃₋₁ = 11 - 1 = 10

    a₅ = a₄₊₁ = a₄₋₁ = 10 - 1 = 9

    a₆ = 2*9 = 18

    We can therefore see that limn→[infinity] does not exist.
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