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Math Help - Sequence

  1. #1
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    Sequence

    In the sequence a_1,a_2,\ldots, a_{80}:
    • a_i>0\,\,\forall 1\le i\le 80
    • a_i=a_{i-1}a_{i+1}\,\,\forall 2\le i\le 79
    • \prod_{i=1}^{40}a_i=8=\prod_{i=1}^{80}a_i

    What can a_1,a_2,\ldots , a_8 be?
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  2. #2
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    Hello james_bond
    Quote Originally Posted by james_bond View Post
    In the sequence a_1,a_2,\ldots, a_{80}:
    • a_i>0\,\,\forall 1\le i\le 80
    • a_i=a_{i-1}a_{i+1}\,\,\forall 2\le i\le 79
    • \prod_{i=1}^{40}a_i=8=\prod_{i=1}^{80}a_i

    What can a_1,a_2,\ldots , a_8 be?
    Re-write the recurrence relation as a_{i+1}=\frac{a_i}{a_{i-1}}, and then look at the first few terms, given a_1 and a_2:

    a_3=\frac{a_2}{a_1}

    a_4=\frac{a_3}{a_2}=\frac{1}{a_1}

    a_5=\frac{a_4}{a_3}=\frac{1}{a_2}

    a_6=\frac{a_5}{a_4}=\frac{a_1}{a_2}

    a_7 = ... = a_1

    a_8 = ... = a_2

    So the sequence repeats after six terms.

    Then note that \prod_{i=1}^6a_i=a_1a_2\cdot\frac{a_2}{a_1}\cdot\f  rac{1}{a_1}\cdot\frac{1}{a_2}\cdot\frac{a_1}{a_2}=  1

    \Rightarrow \prod_{i=1}^{36}a_i=1

    \Rightarrow \prod_{i=1}^{40}a_i=a_1a_2a_3a_4=\frac{{a_2}^2}{a_  1}=8

    \Rightarrow a_1=\frac{{a_2}^2}{8} (1)

    Similarly \prod_{i=1}^{78}a_i=1

    \Rightarrow  \prod_{i=1}^{80}a_i=a_1a_2=8

     \Rightarrow {a_2}^3=64 from (1)

    \Rightarrow a_2=4 and a_1 = 2

    So the sequence is 2, 4, 2, \frac{1}{2}, \frac{1}{4},\frac{1}{2}, 2, 4, ...

    Grandad
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