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Math Help - how to solve recurrence relations?

  1. #1
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    how to solve recurrence relations?

    solve
    i)an+1-2an=5, n>=0, a0=1
    ii)an+1=an+(2n+3), n>=0, a0=1
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  2. #2
    MHF Contributor

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    For the second one, write: a_n=a_0+\sum_{k=0}^{n-1}(a_{k+1}-a_k) (this is a telescopic sum), from which you deduce, using your equation, a_n=1+\sum_{k=0}^{n-1}(2k+3) = 1 + (n-1)n + 3n (remember \sum_{k=0}^n k=\frac{n(n+1)}{2}).

    For the first one, you know a_{n+1}-2a_n, so the telescopic trick doesn't work directly. However, it works for {a_n\over 2^n}: we have \frac{a_{n+1}}{2^{n+1}}-\frac{a_n}{2^n}=\frac{5}{2^{n+1}}. A telescopic summation gives: \frac{a_n}{2^n}=\frac{a_0}{2^0}+\sum_{k=0}^{n-1}\left(\frac{a_{k+1}}{2^{k+1}}-\frac{a_k}{2^k}\right)=1+\sum_{k=0}^{n-1}\frac{5}{2^{k+1}} =1+\frac{5}{2}\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}} (remember the sum of terms in a geometric sequence)

    Laurent.
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