# Solving a recursion relation

1. Nov 25, 2015

### spaghetti3451

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. Nov 25, 2015

### geoffrey159

$\frac{y_k}{y_1} = \prod_{i = 2}^k \frac{y_i}{y_{i-1}}$

3. Nov 27, 2015

### spaghetti3451

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!

4. Nov 27, 2015

### geoffrey159

You're welcome