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If-P-n-denotes-the-product-of-the-binomial-coefficients-in-the-expansion-of-1-x-n-then-P-n-1-P-n-equals-

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

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