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

$$if{(1+x)}^n=\sum _{i=0}^n{a}_i{x}^{i}and$$$${(1+x)}^{n+1}=\sum _{i=0}^{n+1}{b}_i{x}^{i}calculate$$$$\frac{\coprod _{i=0}^n{a}_i}{\prod _{i=0}^{n+1}{b}_i}.$$

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

$$wehave{(x+1)}^n=\sum _{i=0}^n{C}_n^i{x}^i\Rightarrow a_i=C_n^i=\frac{n!}{i!(n-i)!}$$$$also{(x+1)}^{n+1}=\sum _{i=0}^{n+1}{C}_{n+1}^i{x}^i\Rightarrow b_i=C_{n+1}^i$$$$=\frac{(n+1)!}{i!(n+1-i)!}\Rightarrow$$$$\frac{\prod _{i=0}^n{a}_i}{\prod _{i=0}^{n+1}{b}_i}=\frac{\prod _{i=0}^n\frac{n!}{i!(n-i)!}}{\prod _{i=0}^{n+1}\frac{(n+1)!}{i!(n+1-i)!}}$$$$=\frac{1}{n+1}\frac{\prod _{i=0}^n\frac{1}{(n-i)!}}{\prod _{i=0}^n\frac{1}{(n+1-i)!}}=\frac{1}{n+1}\prod _{i=0}^n\frac{(n+1-i)!}{(n-i)!}$$$$=\frac{1}{n+1}\prod _{i=0}^n(n+1-i)$$$$=\frac{1}{n+1}(n+1)n(n-1).....1=n!$$$$$$$$$$

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

$$=\frac{a_0}{b_0}\times \frac{a_1}{b_1}\times \frac{a_2}{b_2}\times ...\times \frac{a_r}{b_r}\times ...\times \frac{a_n}{b_n}\times \frac{1}{b_{n+1}}$$$$now{a}_r={nC}_r=\frac{n!}{r!(n-r)!}$$$$b_r=n+1_{C_r}=\frac{(n+1)!}{r!(n+1-r)!}=\frac{(n+1)n!}{r!(n+1-r)(n-r)!}$$$$so\frac{a_r}{b_r}=\frac{n+1}{n+1-r}$$$$\frac{a_0}{b_0}\times \frac{a_1}{b_1}\times \frac{a_2}{b_2}\times ...\times \frac{a_n}{b_n}\times \frac{1}{b_{n+1}}$$$$=\frac{n+1}{n+1-0}\times \frac{n+1}{n+1-1}\times \frac{n+1}{n+1-2}\times ...\times \frac{n+1}{n+1-n}\times \frac{1}{1}$$$$b_{n+1}=n+1_{C_{n+1}}=1$$$$=\frac{{(n+1)}^{n+1}}{(n+1)!}plscheck$$$$$$

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