Prove-that-n-0-2-n-1-2-n-2-2-n-n-2-2n-n-

Question Number 166110 by naka3546 last updated on 13/Feb/22

Prove that

{(\begin{matrix} n \\ 0 \end{matrix})}^{2} + {(\begin{matrix} n \\ 1 \end{matrix})}^{2} + {(\begin{matrix} n \\ 2 \end{matrix})}^{2} + \dots + {(\begin{matrix} n \\ n \end{matrix})}^{2} = (\begin{matrix} 2 n \\ n \end{matrix})

Answered by qaz last updated on 13/Feb/22

{(\begin{matrix} n \\ 0 \end{matrix})}^{2} + {(\begin{matrix} n \\ 1 \end{matrix})}^{2} + \dots + {(\begin{matrix} n \\ n \end{matrix})}^{2}

= \underset{k = 0}{\sum^{n}} {(\begin{matrix} n \\ k \end{matrix})}^{2}

= \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) (\begin{matrix} n \\ n - k \end{matrix})

= [z^{n}] {(1 + z)}^{n} \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) z^{k}

= [z^{n}] {(1 + z)}^{2 n}

= (\begin{matrix} 2 n \\ n \end{matrix})

Commented by naka3546 last updated on 13/Feb/22

thank you, sir .