Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

Question Number 52600 by Tip Top last updated on 10/Jan/19

If P_n denotes the product of the binomial coefficients in the expansion of (1+x)^n , then (P_(n+1) /P_n ) equals

$$\mathrm{If}\:{P}_{{n}} \:\mathrm{denotes}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{binomial}\: \\ $$$$\mathrm{coefficients}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{1}+{x}\right)^{{n}} , \\ $$$$\mathrm{then}\:\frac{{P}_{{n}+\mathrm{1}} }{{P}_{{n}} }\:\mathrm{equals} \\ $$

Commented by Abdo msup. last updated on 10/Jan/19

we have (x+1)^n =Σ_(k=0) ^n C_n ^k x^k ⇒ P_n =Π_(k=0) ^n C_n ^k =C_n ^o .C_n ^1 .....C_n ^n ⇒ P_(n+1) =C_(n+1) ^0 .C_(n+1) ^1 ...C_(n+1) ^(n+1) ⇒ (P_(n+1) /P_n ) =((Π_(k=0) ^n C_n ^k )/(Π_(k=0) ^(n+1) C_(n+1) ^k )) =((Π_(k=0) ^n C_n ^k )/(Π_(k=0) ^n (((n+1)!)/(k!(n+1−k)!)))) =((Π_(k=0) ^n C_n ^k )/(Π_(k=0) ^n ((n+1)/((n+1−k))) Π_(k=0) ^n ((n!)/(k!(n−k)!)))) =Π_(k=0) ^n (((n+1−k))/(n+1)) = (((n+1)!)/((n+1)^(n+1) )) =(((n+1)n!)/((n+1)(n+1)^n )) =((n!)/((n+1)^n )) .

$${we}\:{have}\:\left({x}+\mathrm{1}\right)^{{n}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\Rightarrow \\ $$$${P}_{{n}} =\prod_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:={C}_{{n}} ^{{o}} \:.{C}_{{n}} ^{\mathrm{1}} \:.....{C}_{{n}} ^{{n}} \:\Rightarrow \\ $$$${P}_{{n}+\mathrm{1}} \:={C}_{{n}+\mathrm{1}} ^{\mathrm{0}} \:.{C}_{{n}+\mathrm{1}} ^{\mathrm{1}} \:\:...{C}_{{n}+\mathrm{1}} ^{{n}+\mathrm{1}} \:\Rightarrow \\ $$$$\frac{{P}_{{n}+\mathrm{1}} }{{P}_{{n}} }\:=\frac{\prod_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} }{\prod_{{k}=\mathrm{0}} ^{{n}+\mathrm{1}} \:{C}_{{n}+\mathrm{1}} ^{{k}} }\:=\frac{\prod_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} }{\prod_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\left({n}+\mathrm{1}\right)!}{{k}!\left({n}+\mathrm{1}−{k}\right)!}} \\ $$$$=\frac{\prod_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} }{\prod_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{{n}+\mathrm{1}}{\left({n}+\mathrm{1}−{k}\right)}\:\:\prod_{{k}=\mathrm{0}} ^{{n}} \:\frac{{n}!}{{k}!\left({n}−{k}\right)!}} \\ $$$$=\prod_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\left({n}+\mathrm{1}−{k}\right)}{{n}+\mathrm{1}}\:=\:\frac{\left({n}+\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:=\frac{\left({n}+\mathrm{1}\right){n}!}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)^{{n}} } \\ $$$$=\frac{{n}!}{\left({n}+\mathrm{1}\right)^{{n}} }\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

p_n =nc_0 ×nc_1 ×nc_2 ...nc_n ←product of nterms p_(n+1) =n+1c_0 ×n+1c_1 ×n+1c_2 ..×n+1c_(n+1) ←product of (n+1)terms now ratio of r+1 th term =((n+1c_r )/(nc_r )) =((((n+1)!)/(r!(n+1−r)!))/((n!)/(r!(n−r)!)))=(((n+1)!)/(n!))×(((n−r)!)/((n+1−r)!)) =((n+1)/(n+1−r)) so (p_(n+1) /p_n ).=(/) =((n+1)/(n+1−0))×((n+1)/(n+1−1))×((n+1)/(n+1−2))... =(((n+1)^n )/((n+1)!))

$${p}_{{n}} ={nc}_{\mathrm{0}} ×{nc}_{\mathrm{1}} ×{nc}_{\mathrm{2}} ...{nc}_{{n}} \:\:\leftarrow{product}\:{of}\:{nterms} \\ $$$${p}_{{n}+\mathrm{1}} ={n}+\mathrm{1}{c}_{\mathrm{0}} ×{n}+\mathrm{1}{c}_{\mathrm{1}} ×{n}+\mathrm{1}{c}_{\mathrm{2}} ..×{n}+\mathrm{1}{c}_{{n}+\mathrm{1}} \leftarrow{product}\:{of}\:\left({n}+\mathrm{1}\right){terms} \\ $$$${now}\:{ratio}\:{of}\:{r}+\mathrm{1}\:{th}\:{term} \\ $$$$ \\ $$$$=\frac{{n}+\mathrm{1}{c}_{{r}} }{{nc}_{{r}} } \\ $$$$=\frac{\frac{\left({n}+\mathrm{1}\right)!}{{r}!\left({n}+\mathrm{1}−{r}\right)!}}{\frac{{n}!}{{r}!\left({n}−{r}\right)!}}=\frac{\left({n}+\mathrm{1}\right)!}{{n}!}×\frac{\left({n}−{r}\right)!}{\left({n}+\mathrm{1}−{r}\right)!} \\ $$$$=\frac{{n}+\mathrm{1}}{{n}+\mathrm{1}−{r}} \\ $$$${so}\:\frac{{p}_{{n}+\mathrm{1}} }{{p}_{{n}} }.=\frac{}{} \\ $$$$=\frac{{n}+\mathrm{1}}{{n}+\mathrm{1}−\mathrm{0}}×\frac{{n}+\mathrm{1}}{{n}+\mathrm{1}−\mathrm{1}}×\frac{{n}+\mathrm{1}}{{n}+\mathrm{1}−\mathrm{2}}... \\ $$$$=\frac{\left({n}+\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)!} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com