prove-that-k-n-1-k-n-n-1-n-pleas-sir-help-me-

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})

Facebook Tweet Pin