# Combinatorial Identity

#### BrownianMan

Let $$\displaystyle B_{k}(x)=\sum_{n\geq 0}\binom{n}{k}\frac{x^{n}}{n!}$$. Show that

$$\displaystyle B_{k}(x)B_{l}(x)=\frac{1}{2^{k+l}}\binom{k+l}{l}B_{k+l}(2x)$$.

I'm having some trouble with this one. Does anyone have any hints?

#### girdav

Use Cauchy product and Chu-Vandermonde equality.

#### BrownianMan

I did that and I got $$\displaystyle B_{k}(x)B_l(x)=\binom{n}{k+l}\frac{(2x)^{n}}{n!}=B_{k+l}(2x)$$.

I'm not sure where the $$\displaystyle \frac{1}{2^{k+l}}\binom{k+l}{l}$$ comes from.

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Anyone?