Show that Σ_(k=0) ^n (C_n ^k )^2 =C


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Question Number 207374 by sniper237 last updated on 13/May/24

$S h o w t h a t \underset{k = 0}{\sum^{n}} {(C_{n}^{k})}^{2} = C_{2 n}^{n}$
	Answered by mr W last updated on 13/May/24

	${(1 + x)}^{n} {(1 + x)}^{n} = {(1 + x)}^{2 n}$ $(\underset{k = 0}{\sum^{n}} C_{k}^{n} x^{k}) (\underset{s = 0}{\sum^{n}} C_{s}^{n} x^{s}) = \underset{k = 0}{\sum^{2 n}} C_{r}^{2 n} x^{r}$ $c o e f . o f x^{n} :$ $k + s = n, r = n$ $\underset{k = 0}{\sum^{n}} C_{k}^{n} C_{s}^{n} = C_{n}^{2 n}$ $\underset{k = 0}{\sum^{n}} C_{k}^{n} C_{n - k}^{n} = C_{n}^{2 n}$ $\Rightarrow \underset{k = 0}{\sum^{n}} {(C_{k}^{n})}^{2} = C_{n}^{2 n}$
	Answered by mr W last updated on 13/May/24

	$a n o t h e r m e t h o d$ $s a y t h e r e a r e n m e n a n d n w o m e n .$ $t o s e l e c t n p e r s o n s f r o m t h e m t h e r e$ $a r e C_{n}^{2 n} w a y s . w e c a n a l s o s e l e c t$ $k m e n a n d n - k w o m e n, t h e r e a r e$ $C_{k}^{n} C_{n - k}^{n} = {(C_{k}^{n})}^{2} w a y s . k c a n b e f r o m$ $0 t o n, t h e r e f o r e t o t a l l y \underset{k = 0}{\sum^{n}} {(C_{k}^{n})}^{2} w a y s .$ $\Rightarrow \underset{k = 0}{\sum^{n}} {(C_{k}^{n})}^{2} = C_{n}^{2 n}$
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