Solving a recursion relation

Question

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.

Answer 1

Multiplying the both sides of $$y_k=k(2j-k+1)y_{k-1}$$ by $\frac{(2j-k)!}{k!}$ gives $$\frac{(2j-k)!}{k!}y_k=\frac{(2j-k+1)!}{(k-1)!}y_{k-1},$$ i.e. $$z_k=z_{k-1}$$ where $$z_k=\frac{(2j-k)!}{k!}y_k$$

So, we have $$\frac{(2j-k)!}{k!}y_k=z_k=z_{k-1}=\cdots =z_1=z_0=\frac{(2j-0)!}{0!}y_0=(2j)!,$$ i.e. $$y_k=\frac{k!(2j)!}{(2j-k)!}$$

Answer 2

I forgot to mention that $y_{0}=1$.

All right, then, we have

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

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

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

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

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