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Question Number 39520 by math khazana by abdo last updated on 07/Jul/18
if (1+x)^n =Σ_(i=0) ^n a_i x^i and (1+x)^(n+1) =Σ_(i=0) ^(n+1) b_i x^i calculate ((∐_(i=0) ^n a_i )/(Π_(i=0) ^(n+1) b_i )) .
$${if}\:\left(\mathrm{1}+{x}\right)^{{n}} \:=\sum_{{i}=\mathrm{0}} ^{{n}} \:{a}_{{i}} {x}^{{i}} \:\:\:\:{and} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{1}} \:=\sum_{{i}=\mathrm{0}} ^{{n}+\mathrm{1}} \:{b}_{{i}} \:{x}^{{i}} \:\:{calculate} \\ $$$$\frac{\coprod_{{i}=\mathrm{0}} ^{{n}} \:{a}_{{i}} }{\prod_{{i}=\mathrm{0}} ^{{n}+\mathrm{1}} \:{b}_{{i}} }\:. \\ $$
Commented by math khazana by abdo last updated on 07/Jul/18
we have (x+1)^n =Σ_(i=0) ^n C_n ^i x^i ⇒a_i =C_n ^i =((n!)/(i!(n−i)!)) also (x+1)^(n+1) =Σ_(i=0) ^(n+1) C_(n+1) ^i x^i ⇒ b_i = C_(n+1) ^i =(((n+1)!)/(i!(n+1−i)!)) ⇒ ((Π_(i=0) ^n a_i )/(Π_(i=0) ^(n+1) b_i )) = ((Π_(i=0) ^n ((n!)/(i!(n−i)!)))/(Π_(i=0) ^(n+1) (((n+1)!)/(i!(n+1−i)!)))) = (1/(n+1)) ((Π_(i=0) ^n (1/((n−i)!)))/(Π_(i=0) ^n (1/((n+1−i)!)))) =(1/(n+1)) Π_(i=0) ^n (((n+1−i)!)/((n−i)!)) =(1/(n+1))Π_(i=0) ^n (n+1−i) = (1/(n+1)) (n+1)n(n−1).....1 =n!
$${we}\:{have}\:\left({x}+\mathrm{1}\right)^{{n}} \:=\sum_{{i}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{i}} \:{x}^{{i}} \:\Rightarrow{a}_{{i}} ={C}_{{n}} ^{{i}} \:=\frac{{n}!}{{i}!\left({n}−{i}\right)!} \\ $$$${also}\:\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} \:=\sum_{{i}=\mathrm{0}} ^{{n}+\mathrm{1}} \:{C}_{{n}+\mathrm{1}} ^{{i}} \:{x}^{{i}} \:\Rightarrow\:{b}_{{i}} =\:{C}_{{n}+\mathrm{1}} ^{{i}} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)!}{{i}!\left({n}+\mathrm{1}−{i}\right)!}\:\Rightarrow \\ $$$$\frac{\prod_{{i}=\mathrm{0}} ^{{n}} \:{a}_{{i}} }{\prod_{{i}=\mathrm{0}} ^{{n}+\mathrm{1}} \:{b}_{{i}} }\:=\:\frac{\prod_{{i}=\mathrm{0}} ^{{n}} \:\:\:\:\frac{{n}!}{{i}!\left({n}−{i}\right)!}}{\prod_{{i}=\mathrm{0}} ^{{n}+\mathrm{1}} \:\:\:\frac{\left({n}+\mathrm{1}\right)!}{{i}!\left({n}+\mathrm{1}−{i}\right)!}} \\ $$$$=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\frac{\prod_{{i}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({n}−{i}\right)!}}{\prod_{{i}=\mathrm{0}} ^{{n}} \frac{\mathrm{1}}{\left({n}+\mathrm{1}−{i}\right)!}}\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\:\prod_{{i}=\mathrm{0}} ^{{n}} \:\frac{\left({n}+\mathrm{1}−{i}\right)!}{\left({n}−{i}\right)!} \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\prod_{{i}=\mathrm{0}} ^{{n}} \:\left({n}+\mathrm{1}−{i}\right) \\ $$$$=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\left({n}+\mathrm{1}\right){n}\left({n}−\mathrm{1}\right)…..\mathrm{1}\:\:={n}! \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18
=(a_0 /b_0 )×(a_1 /b_1 )×(a_2 /b_2 )×...×(a_r /b_r )×...×(a_n /b_n )×(1/b_(n+1) ) now a_r =nC_r =((n!)/(r!(n−r)!)) b_r =n+1_C_r =(((n+1)!)/(r!(n+1−r)!))=(((n+1)n!)/(r!(n+1−r)(n−r)!)) so (a_r /b_r )=((n+1)/(n+1−r)) (a_0 /b_0 )×(a_1 /b_1 )×(a_2 /b_2 )×...×(a_n /b_n )×(1/b_(n+1) ) =((n+1)/(n+1−0))×((n+1)/(n+1−1))×((n+1)/(n+1−2))×...×((n+1)/(n+1−n))×(1/1) b_(n+1) =n+1_C_(n+1) =1 =(((n+1)^(n+1) )/((n+1)!)) pls check
$$=\frac{{a}_{\mathrm{0}} }{{b}_{\mathrm{0}} }×\frac{{a}_{\mathrm{1}} }{{b}_{\mathrm{1}} }×\frac{{a}_{\mathrm{2}} }{{b}_{\mathrm{2}} }×…×\frac{{a}_{{r}} }{{b}_{{r}} }×…×\frac{{a}_{{n}} }{{b}_{{n}} }×\frac{\mathrm{1}}{{b}_{{n}+\mathrm{1}} } \\ $$$${now}\:{a}_{{r}} ={nC}_{{r}} =\frac{{n}!}{{r}!\left({n}−{r}\right)!} \\ $$$${b}_{{r}} ={n}+\mathrm{1}_{{C}_{{r}} } =\frac{\left({n}+\mathrm{1}\right)!}{{r}!\left({n}+\mathrm{1}−{r}\right)!}=\frac{\left({n}+\mathrm{1}\right){n}!}{{r}!\left({n}+\mathrm{1}−{r}\right)\left({n}−{r}\right)!} \\ $$$${so}\:\frac{{a}_{{r}} }{{b}_{{r}} }=\frac{{n}+\mathrm{1}}{{n}+\mathrm{1}−{r}} \\ $$$$\frac{{a}_{\mathrm{0}} }{{b}_{\mathrm{0}} }×\frac{{a}_{\mathrm{1}} }{{b}_{\mathrm{1}} }×\frac{{a}_{\mathrm{2}} }{{b}_{\mathrm{2}} }×…×\frac{{a}_{{n}} }{{b}_{{n}} }×\frac{\mathrm{1}}{{b}_{{n}+\mathrm{1}} } \\ $$$$=\frac{{n}+\mathrm{1}}{{n}+\mathrm{1}−\mathrm{0}}×\frac{{n}+\mathrm{1}}{{n}+\mathrm{1}−\mathrm{1}}×\frac{{n}+\mathrm{1}}{{n}+\mathrm{1}−\mathrm{2}}×…×\frac{{n}+\mathrm{1}}{{n}+\mathrm{1}−{n}}×\frac{\mathrm{1}}{\mathrm{1}} \\ $$$${b}_{{n}+\mathrm{1}} ={n}+\mathrm{1}_{{C}_{{n}+\mathrm{1}} } =\mathrm{1} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)!}\:\:{pls}\:{check} \\ $$$$ \\ $$

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