prove that ( _( k−1) ^(2k) ) =^? Σ_(r=0) ^(k−1) ( _( r) ^( k) ) (^( )


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Question Number 164555 by mnjuly1970 last updated on 18/Jan/22

$p r o v e t h a t$ $Extra \left or missing \right$ $- - - -$
	Answered by mr W last updated on 19/Jan/22

	${(1 + x)}^{2 k} = {(1 + x)}^{k} {(1 + x)}^{k}$ $\underset{q = 0}{\sum^{2 k}} (\begin{matrix} 2 k \\ q \end{matrix}) x^{q} = [\underset{r = 0}{\sum^{k}} (\begin{matrix} k \\ r \end{matrix}) x^{r}] [\underset{p = 0}{\sum^{k}} (\begin{matrix} k \\ p \end{matrix}) x^{p}]$ ${l e t}^{'} s s e e t e r m x^{q} w i t h q = k - 1$ $r + p = k - 1$ $\Rightarrow p = k - 1 - r ⩾ 0 \Rightarrow r ⩽ k - 1$ $(\begin{matrix} 2 k \\ k - 1 \end{matrix}) = \underset{r = 0}{\sum^{k - 1}} (\begin{matrix} k \\ r \end{matrix}) (\begin{matrix} k \\ k - r - 1 \end{matrix})$ $(\begin{matrix} 2 k \\ k - 1 \end{matrix}) = \underset{r = 0}{\sum^{k - 1}} (\begin{matrix} k \\ r \end{matrix}) (\begin{matrix} k \\ r + 1 \end{matrix}) ✓$
	Commented by mnjuly1970 last updated on 19/Jan/22

	$e x c e l l e n t s i r W$ $g r a t e f u l . . .$
	Answered by hmrsh last updated on 18/Jan/22

	$First of all, I would like to review$ $a basic combinatorial identity .$ $∙ (\begin{matrix} n \\ k \end{matrix}) = (\begin{matrix} n \\ n - k \end{matrix})$ $Well, {It}^{'} s interesting to point out$ $that the question is relevant to a$ $famous combinatorial identity, {that}^{'} s$ $called ‘ ‘ {Vandermonde}^{'} s {Identity}^{″} .$ $* \underset{k = 0}{\sum^{r}} (\begin{matrix} m \\ k \end{matrix}) (\begin{matrix} n \\ r - k \end{matrix}) = (\begin{matrix} m + n \\ r \end{matrix})$ $If we replace n with r and m, we have :$ $∙ \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) (\begin{matrix} n \\ n - k \end{matrix}) = \underset{k = 0}{\sum^{n}} {(\begin{matrix} n \\ k \end{matrix})}^{2} = (\begin{matrix} 2 n \\ n \end{matrix})$ $Now {let}^{'} s prove the {Vandermonde}^{'} s$ $identity with combinatorial reasons .$ $Note that {it}^{'} s easier to use combinatorial$ $reason than to use induction or expand$ $the equation .$ $Just enough to use your creativity$ $and make a counting problem that$ $both LHS and RHS of the identity$ $applied that problem .$ $Like this for prove * :$ $‘ ‘ We want to form a r - member team$ $from m men and n women .$ $In how many ways we can make that team ?^{″}$ $Note that the RHS of * is directly$ $the answer .$ $How about the LHS ?$ $You can count the ways with moderate$ $the number of men .$ $e . g no men is in team, just 1 man$ $is in team and so on .$ $Summarize the different states$ $with sigma .$ $Now your Question :$ $\underset{r = 0}{\sum^{k - 1}} (\begin{matrix} k \\ r \end{matrix}) (\begin{matrix} k \\ r + 1 \end{matrix}) = \underset{r = 0}{\sum^{k - 1}} (\begin{matrix} k \\ r \end{matrix}) (\begin{matrix} k \\ (k - 1) - r \end{matrix}) = (\begin{matrix} 2 k \\ k - 1 \end{matrix})$ $Of course you can prove your equation$ $without using the {Vandermonde}^{'} s$ $identity . Just use the proof which$ $we say in terms that solves your problem .$
	Commented by mnjuly1970 last updated on 19/Jan/22

	$t h a n k s a l o t s i r$
	Commented by Rasheed.Sindhi last updated on 19/Jan/22

	$W i t h g o o d e x p l a n a t i o n!$
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