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Solving a recursion relation

  1. Nov 25, 2015 #1
    I have the recursion relation ##y_{k}=k(2j-k+1)y_{k-1}##

    and I would like to solve it to obtain ##y_{k}=\frac{k!(2j)!}{(2j-k)!}##.

    Can you provide some hints on how I might proceed?

    P.S.: ##j## is a constant.
     
  2. jcsd
  3. Nov 25, 2015 #2
    ## \frac{y_k}{y_1} = \prod_{i = 2}^k \frac{y_i}{y_{i-1}} ##
     
  4. Nov 27, 2015 #3
    I forgot to mention that ##y_{0}=1##.

    All right, then, we have

    ##\frac{y_k}{y_0} = \prod_{i = 1}^k \frac{y_i}{y_{i-1}}##

    ##y_k = \prod_{i=1}^{k} i(2j-i+1)##

    ##y_k = \bigg(\prod_{i=1}^{k} i\bigg) \bigg(\prod_{i=1}^{k} (2j-i+1)\bigg)##

    ##y_k = (k!) \bigg(\prod_{i=1}^{k} (2j-(i-1))\bigg)##

    ##y_k = (k!) \frac{(2j)!}{(2j-k)!}##

    Thanks!
     
  5. Nov 27, 2015 #4
    You're welcome
     
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