# Evaluate sum_{k=0}^{n} ((2k),(k)) ((2n-2k),(n-k))

Braylon Lester 2022-07-15 Answered
Discrete math evaluate
Evaluate $\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{2k}{k}\right)\left(\genfrac{}{}{0}{}{2n-2k}{n-k}\right)$
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Tolamaes04
Step 1
Cauchy product
Note, the binomial expression in OPs question is a Cauchy product which we get as the coefficient of the multiplication of two power series. In case both series are equal, say $A\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_{k}{x}^{k}$ we obtain
$\begin{array}{}\text{(1)}& {A}^{2}\left(x\right)={\left(\sum _{n\ge 0}{a}_{n}{x}^{n}\right)}^{2}=\sum _{n\ge 0}\left(\sum _{k=0}^{n}{a}_{k}{a}_{n-k}\right){x}^{n}\end{array}$
Applying (1) and using the hint stated in OPs question with ${a}_{n}=\left(\genfrac{}{}{0}{}{2n}{n}\right)$ we get
$\begin{array}{rl}\sum _{n\ge 0}& \left(\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{2k}{k}\right)\left(\genfrac{}{}{0}{}{2n-2k}{n-k}\right)\right){x}^{n}\\ & ={\left(\sum _{n\ge 0}\left(\genfrac{}{}{0}{}{2n}{n}\right){x}^{n}\right)}^{2}\\ & ={\left(\frac{1}{\sqrt{1-4x}}\right)}^{2}\\ & =\frac{1}{1-4x}\\ & =\sum _{n\ge 0}{4}^{n}{x}^{n}\end{array}$
Step 2
We conclude by comparing coefficients
$\begin{array}{r}\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{2k}{k}\right)\left(\genfrac{}{}{0}{}{2n-2k}{n-k}\right)={4}^{n}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}n\ge 0\end{array}$

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Dean Summers
Step 1
Coefficient of - operator
Use the coefficient of operator $\left[{x}^{n}\right]$ to denote the coefficient of ${x}^{n}$ of a series $A\left(x\right)=\sum _{k\ge 0}{a}_{k}{x}^{k}$
Step 2
We obtain for $n\ge 0$
$\begin{array}{rl}\sum _{k=0}^{n}& \left(\genfrac{}{}{0}{}{2k}{k}\right)\left(\genfrac{}{}{0}{}{2n-2k}{n-k}\right)\\ & =\sum _{k=0}^{n}\left[{x}^{k}\right]\frac{1}{\sqrt{1-4x}}\left[{y}^{n-k}\right]\frac{1}{\sqrt{1-4y}}\\ \text{(2)}& & =\left[{y}^{n}\right]\frac{1}{\sqrt{1-4y}}\sum _{k=0}^{n}{y}^{k}\left[{x}^{k}\right]\frac{1}{\sqrt{1-4x}}\text{(3)}& & =\left[{y}^{n}\right]\frac{1}{\sqrt{1-4y}}\frac{1}{\sqrt{1-4y}}& =\left[{y}^{n}\right]\frac{1}{1-4y}\\ & =\left[{y}^{n}\right]\sum _{n\ge 0}{4}^{n}{y}^{n}\\ & ={4}^{n}\end{array}$

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