Question

combinatrics

Answer 1

\[\Huge \sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)^2=\left(\begin{matrix}2n \\ n\end{matrix}\right)\]

Answer 2

prove

Answer 3

\[\left(\begin{matrix}n \\ 0\end{matrix}\right)^2+\left(\begin{matrix}n \\ 1\end{matrix}\right)^2+...+\left(\begin{matrix}n \\ n\end{matrix}\right)^2\]

Answer 4

so i started also like this
\[\large \sum_{i=0}^{j}\left(\begin{matrix}r+i-1 \\ r-1\end{matrix}\right)=\left(\begin{matrix}r+j \\ j\end{matrix}\right)\]
using this identity let j=r
\[\large \sum_{i=0}^{r}\left(\begin{matrix}r+i-1 \\ r-1\end{matrix}\right)=\left(\begin{matrix}2r \\ r\end{matrix}\right)\]

Answer 5

i wanna make iteration of this sum to make it of the form
\[\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)^2\]

Answer 6

what are the coefficients of x^n when
we have this product
\[\xi(x)=\large \sum_{i=0}^{n}\left(\begin{matrix}n \\ i\end{matrix}\right)x^{n-i}\dot {} \sum_{i=0}^{n}\left(\begin{matrix}n \\ i\end{matrix}\right)x^i\]

Answer 7

i think this is it,the coeffecients are our sum
\[[\color{blue}{\left(\begin{matrix}n \\ 0\end{matrix}\right)+\left(\begin{matrix}n \\ 1\end{matrix}\right)x+\left(\begin{matrix}n \\ 2\end{matrix}\right)x^2+...\left(\begin{matrix}n \\ n\end{matrix}\right)x^n} ][\color{red}{\left(\begin{matrix}n \\ 0\end{matrix}\right)x^n+\left(\begin{matrix}n \\ 1\end{matrix}\right)x^{n-1}+\left(\begin{matrix}n \\ 2\end{matrix}\right)x^{n-2}+...\left(\begin{matrix}n \\ n\end{matrix}\right)x}]\]
coe effeicients are there fore
\[\left(\begin{matrix}n \\ 0\end{matrix}\right)^2+\left(\begin{matrix}n \\ 1\end{matrix}\right)^2+\left(\begin{matrix}n \\ 2\end{matrix}\right)^2+...\left(\begin{matrix}n \\ n\end{matrix}\right)^2\]
\[\huge \xi (x)=(1+x)^n(1+x)^n=\\ \huge (1+x)^{2n}=\sum_{i=0}^{2n}\left(\begin{matrix}2n \\ i\end{matrix}\right)x^{2n}\]