Showing $\sum_{\alpha=0}^k \binom{k+15}{\alpha + 10}\binom{k}{\alpha}=\frac{2^{2k+15}(k+6)(k+7)\left(\frac{2k+15}{2}\right)!}{\sqrtπ (k+10)!}$

Question

I want to simplify $\sum_{\alpha=0}^k \binom{k+15}{\alpha + 10}\binom{k}{\alpha}$ but this can't be directly simplified using Vandermonde Chu Identity. Wolfram shows a nice closed form expression but I can't get to it. My attempt $\sum_{k=0}^n \binom{k+15}{5+k-\alpha}\binom{k}{\alpha}=\binom{2k+15}{5+k}$ by thinking of this sum as choosing a set of 5+k objects from 2k+15 objects in total but I feel that this isn't correct as some terms are missing

Answer 1

Prove that $$\binom{2k+15}{k+5}=\frac{2^{2k+15}(k+6)(k+7)\left(\frac{2k+15}{2}\right)!}{\sqrtπ (k+10)!}$$

Number of $2$'s in $\left(\frac{2k+15}{2}\right)!=\frac{2k+15+1}{2}=k+8$. So, the RHS becomes \begin{align*} &\Rightarrow\frac{2^{k+7}(k+6)(k+7)(2k+15)!!\ \Gamma\left(\frac12\right)}{\sqrtπ(k+10)!}\\ &=\frac{2^{k+5}(k+5)!(2k+12)(2k+14)(2k+15)!!}{(k+10)!(k+5)!}&\left(\because \Gamma\left(\frac12\right)=\sqrt{\pi}\right)\\ &=\frac{(2k+10)!!(2k+12)(2k+14)}{(k+10)!(k+5)!}\frac{(2k+15)!}{(2k+14)!!}\\ &=\frac{(2k+15)!}{(k+10)!(k+5)!}\\ \end{align*}

PS: I think that Wolfram Alpha's algorithm doesn't apply the common logic of choosing $k+5$ people from $2k+15$.

Answer 2

$$S=\sum_{k=0}^n {n+15 \choose k+10} {n\choose k}$$ $${n+15 \choose k+10}= \text{Co-efficient of} ~~x^{k+10}~~ in ~~(1+x)^{n+15}$$ Multiply by ${n choose k}$ both sides and sum over $k$, theb $$S=\sum_{k=0}^{n}{n+15 \choose k+10} {n\choose k}=\text{Co-efficient of} ~~x^{10}~~in~~(1+x)^{n+15} \sum_{k=0}^n {n \choose k} x^{-k}$$ $$\implies S=\text{Co-efficient of }~~x^{10}~~ in ~~ (1+x)^{n+15}(1+1/x)^n$$ $$\implies S=\text{Co-efficient of }~~x^{n+10}~~ in ~~ (1+x)^{n+15}(1+x)^n ={2n+15 \choose n+10}$$ Next, $$S={n+15 \choose n+10}=\frac{(2n+15)!}{(n+5)! (n+10)!}=\frac{\Gamma(2n+16)}{(n+5)! (n+10)!}$$ Using the property of Gamma function $$\Gamma(2z)=(2\pi)^{-1/2}~~2^{2z-1/2}~\Gamma(z)~~ \Gamma(z+1/2)$$ We can wrote $$S=(2\pi)^{-1/2}~ 2^{2n+16-1/2}~ \frac{\Gamma(n+8) \Gamma(n+8+1/2)}{(n+5)! ~(n+10)!}=(2\pi)^{-1/2}~2^{2n+16-1/2} \frac{(n+7)!~ (\frac{2n+15}{2})!}{(n+5)! ~(n+10)!}$$ $$\implies 2^{2n+15}~ \frac{(n+6)(n+7)~ (\frac{2n+15}{2})!}{\sqrt{\pi}~(n+10)!}$$