Prove-that-1-n-r-n-1-r-n-1-r-1-2-n-r-n-r-1-n-1-r-3-n-0-n-1-n-2-n-n-2-n-

Question Number 54740 by gunawan last updated on 10/Feb/19

P r o v e t h a t

1 . (\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n - 1 \\ r \end{matrix}) + (\begin{matrix} n - 1 \\ r - 1 \end{matrix})

2 . (\begin{matrix} n \\ r \end{matrix}) + (\begin{matrix} n \\ r - 1 \end{matrix}) = (\begin{matrix} n + 1 \\ r \end{matrix})

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

Answered by Kunal12588 last updated on 10/Feb/19

3 .^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + \dots +^{n} C_{n} = 2^{n} {e q}^{n} 1

f r o m b i n o m i a l e x p a n s i o n

{(a + b)}^{p} =^{p} C_{0} a^{p} +^{p} C_{1} a^{p - 1} b^{1} + \dots +^{p} C_{r} a^{p - r} b^{r} + \dots +^{p} C_{p} b^{p} [r < p]

c o m p a r i n g w i t h t h e q u e s t i o n

∵ 1^{a n y t h i n g} = 1

{e q}^{n} 1 c a n b e w r i t t e n a s :

^{n} C_{0} 1^{n} +^{n} C_{1} 1^{n - 1} 1^{1} +^{n} C_{2} 1^{n - 2} 1^{2} + \dots +^{n} C_{n} 1^{n}

= {(1 + 1)}^{n} = 2^{n} p r o v e d

Commented by gunawan last updated on 10/Feb/19

nice solution

thank you Sir

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19

a n o t h e r w a y {(1 + x)}^{n} = {n c}_{0} + {n c}_{1} x + \dots + {n c}_{n} x^{n}

p u t x = 1

2^{n} = {n c}_{0} + {n c}_{1} + \dots + {n c}_{n}

Commented by maxmathsup by imad last updated on 11/Feb/19

3) l e t p (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{k} = {(x + 1)}^{n}

x = 1 \Rightarrow \sum_{k = 0}^{n} C_{n}^{k} = 2^{n} \Rightarrow C_{n}^{0} + C_{n}^{1} + C_{n}^{2} + \dots . + C_{n}^{n} = 2^{n} .

Answered by Kunal12588 last updated on 10/Feb/19

^{n} C_{r} =^{n - 1} C_{r} +^{n - 1} C_{r - 1}

1 . R H S

^{n - 1} C_{r} +^{n - 1} C_{r - 1} = \frac{(n - 1)!}{r! (n - 1 - r)!} + \frac{(n - 1)!}{(r - 1)! (n - r)!}

= \frac{(n - 1)!}{r (r - 1)! (n - 1 - r)!} + \frac{(n - 1)!}{(r - 1)! (n - r) (n - 1 - r)!}

= \frac{(n - 1)!}{(r - 1)! (n - 1 - r)!} (\frac{1}{r} + \frac{1}{n - r})

= \frac{(n - 1)!}{(r - 1)! (n - 1 - r)!} (\frac{n}{r (n - r)})

= \frac{n (n - 1)!}{r (r - 1)! (n - r) (n - 1 - r!)} = \frac{n!}{r! (n - r)!}

=^{n} C_{r} = L H S p r o v e d

Answered by Kunal12588 last updated on 10/Feb/19

2 .^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}

L H S

= \frac{n!}{r! (n - r)!} + \frac{n!}{(r - 1)! (n - r + 1)!}

= \frac{n!}{(r - 1)! (n - r)!} (\frac{1}{r} + \frac{1}{n - r + 1})

= \frac{(n + 1)!}{r! (n + 1 - r)!} =^{n + 1} C_{r} = R H S p r o v e d

Commented by Kunal12588 last updated on 10/Feb/19

s a m e a s (1) j u s t r e p l a c e (n - 1) w i t h n