if-1-x-n-i-0-n-a-i-x-i-and-1-x-n-1-i-0-n-1-b-i-x-i-calculate-i-0-n-a-i-i-0-n-1-b-i-

Question Number 39520 by math khazana by abdo last updated on 07/Jul/18

i f {(1 + x)}^{n} = \sum_{i = 0}^{n} a_{i} x^{i} a n d

{(1 + x)}^{n + 1} = \sum_{i = 0}^{n + 1} b_{i} x^{i} c a l c u l a t e

\frac{∐_{i = 0}^{n} a_{i}}{\prod_{i = 0}^{n + 1} b_{i}} .

Commented by math khazana by abdo last updated on 07/Jul/18

w e h a v e {(x + 1)}^{n} = \sum_{i = 0}^{n} C_{n}^{i} x^{i} \Rightarrow a_{i} = C_{n}^{i} = \frac{n!}{i! (n - i)!}

a l s o {(x + 1)}^{n + 1} = \sum_{i = 0}^{n + 1} C_{n + 1}^{i} x^{i} \Rightarrow b_{i} = C_{n + 1}^{i}

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

\frac{\prod_{i = 0}^{n} a_{i}}{\prod_{i = 0}^{n + 1} b_{i}} = \frac{\prod_{i = 0}^{n} \frac{n!}{i! (n - i)!}}{\prod_{i = 0}^{n + 1} \frac{(n + 1)!}{i! (n + 1 - i)!}}

= \frac{1}{n + 1} \frac{\prod_{i = 0}^{n} \frac{1}{(n - i)!}}{\prod_{i = 0}^{n} \frac{1}{(n + 1 - i)!}} = \frac{1}{n + 1} \prod_{i = 0}^{n} \frac{(n + 1 - i)!}{(n - i)!}

= \frac{1}{n + 1} \prod_{i = 0}^{n} (n + 1 - i)

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

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18

= \frac{a_{0}}{b_{0}} \times \frac{a_{1}}{b_{1}} \times \frac{a_{2}}{b_{2}} \times \dots \times \frac{a_{r}}{b_{r}} \times \dots \times \frac{a_{n}}{b_{n}} \times \frac{1}{b_{n + 1}}

n o w a_{r} = {n C}_{r} = \frac{n!}{r! (n - r)!}

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

s o \frac{a_{r}}{b_{r}} = \frac{n + 1}{n + 1 - r}

\frac{a_{0}}{b_{0}} \times \frac{a_{1}}{b_{1}} \times \frac{a_{2}}{b_{2}} \times \dots \times \frac{a_{n}}{b_{n}} \times \frac{1}{b_{n + 1}}

= \frac{n + 1}{n + 1 - 0} \times \frac{n + 1}{n + 1 - 1} \times \frac{n + 1}{n + 1 - 2} \times \dots \times \frac{n + 1}{n + 1 - n} \times \frac{1}{1}

b_{n + 1} = n + 1_{C_{n + 1}} = 1

= \frac{{(n + 1)}^{n + 1}}{(n + 1)!} p l s c h e c k