prove that (_k ^(n+1) )=(_k ^n )+(_(n−1) ^n ) pleas sir help me ?


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Question Number 73590 by mhmd last updated on 13/Nov/19

$p r o v e t h a t (_{k}^{n + 1}) = (_{k}^{n}) + (_{n - 1}^{n})$ $p l e a s s i r h e l p m e ?$
Commented by Cmr 237 last updated on 13/Nov/19

$i t i s$ $(\begin{matrix} n \\ k \end{matrix}) = (\begin{matrix} n - 1 \\ k - 1 \end{matrix}) + (\begin{matrix} n - 1 \\ k \end{matrix})$
Commented by mathmax by abdo last updated on 13/Nov/19

$t h e c o r r e c t f o r m u l a i s C_{n}^{k} = C_{n - 1}^{k} + C_{n - 1}^{k - 1} f o r 1 ⩽ k ⩽ n$ $l e t p r o v e i t w e h a v e$ $C_{n - 1}^{k} + C_{n - 1}^{k - 1} = \frac{(n - 1)!}{k! (n - 1 - k)!} + \frac{(n - 1)!}{(k - 1)! (n - k)!}$ $= \frac{n! (n - k)}{n (k!) (n - k)!} + \frac{n! k}{n k! (n - k)!}$ $= \frac{n - k}{n} C_{n}^{k} + \frac{k}{n} C_{n}^{k} = (1 - \frac{k}{n} + \frac{k}{n}) C_{n}^{k} = C_{n}^{k}$
	Answered by Cmr 237 last updated on 13/Nov/19

	$i f y o u t a k e k = 1 a n d n = 2$ $y o u r f o r m u l w a s n o t t r u e!!!$
	Answered by malwaan last updated on 14/Nov/19

	$(\begin{matrix} n \\ k \end{matrix}) + (\begin{matrix} n \\ k - 1 \end{matrix}) =$ $\frac{n!}{k! (n - k)!} + \frac{n!}{(k - 1)! (n - k + 1)!}$ $= \frac{n!}{k (k - 1)! (n - k)!} +$ $\frac{n!}{(k - 1)! (n - k + 1) (n - k)!}$ $= \frac{n! [(n - k + 1) + k]}{k (k - 1)! (n - k + 1) (n - k)!}$ $= \frac{n! (n + 1)}{k! (n - k + 1)!} = \frac{(n + 1)!}{k! [(n + 1) - k]!}$ $= (\begin{matrix} n + 1 \\ k \end{matrix})$
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