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Question Number 120044 by john santu last updated on 28/Oct/20
Given a_(n+1) = ((2a_n )/((2n+1)(2n+2))) find a_n .
$${Given}\:{a}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{2}{a}_{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)} \\ $$$${find}\:{a}_{{n}} . \\ $$
Answered by Olaf last updated on 29/Oct/20
Let v_k = (a_(k+1) /a_k ) = (2/((2k+1)(2k+2))) Π_(k=0) ^n v_k = Π_(k=0) ^n (a_(k+1) /a_k ) = (a_(n+1) /a_0 ) (telescopic product) and Π_(k=0) ^n v_k = Π_(k=0) ^n (2/((2k+1)(2k+2))) Π_(k=0) ^n v_k = (2^(n+1) /((1.2)(3.4)(5.6)...(2n+1)(2n+2))) Π_(k=0) ^n v_k = (2^(n+1) /((2n+2)!)) ⇒ a_(n+1) = a_0 (2^(n+1) /((2n+2)!)) and a_n = a_0 (2^n /((2n)!))
$$\mathrm{Let}\:{v}_{{k}} \:=\:\frac{{a}_{{k}+\mathrm{1}} }{{a}_{{k}} }\:=\:\frac{\mathrm{2}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{2}\right)} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}{v}_{{k}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\frac{{a}_{{k}+\mathrm{1}} }{{a}_{{k}} }\:=\:\frac{{a}_{{n}+\mathrm{1}} }{{a}_{\mathrm{0}} } \\ $$$$\left(\mathrm{telescopic}\:\mathrm{product}\right) \\ $$$$\mathrm{and}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}{v}_{{k}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\frac{\mathrm{2}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{2}\right)} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}{v}_{{k}} \:=\:\frac{\mathrm{2}^{{n}+\mathrm{1}} }{\left(\mathrm{1}.\mathrm{2}\right)\left(\mathrm{3}.\mathrm{4}\right)\left(\mathrm{5}.\mathrm{6}\right)…\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}{v}_{{k}} \:=\:\frac{\mathrm{2}^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)!} \\ $$$$\Rightarrow\:{a}_{{n}+\mathrm{1}} \:=\:{a}_{\mathrm{0}} \frac{\mathrm{2}^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)!} \\ $$$$\mathrm{and}\:{a}_{{n}} \:=\:{a}_{\mathrm{0}} \frac{\mathrm{2}^{{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$ \\ $$
Commented by bramlexs22 last updated on 29/Oct/20
what the value of a_0 sir?
$${what}\:{the}\:{value}\:{of}\:{a}_{\mathrm{0}} \:{sir}? \\ $$
Commented by Olaf last updated on 29/Oct/20
It′s impossible to know. It is not given in the problem. a_0 is simply the first term in the sequence.
$$\mathrm{It}'\mathrm{s}\:\mathrm{impossible}\:\mathrm{to}\:\mathrm{know}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{not}\:\mathrm{given}\:\mathrm{in}\:\mathrm{the}\:\mathrm{problem}. \\ $$$${a}_{\mathrm{0}} \:\mathrm{is}\:\mathrm{simply}\:\mathrm{the}\:\mathrm{first}\:\mathrm{term}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{sequence}. \\ $$
Answered by Bird last updated on 29/Oct/20
(a_(n+1) /a_n )=(1/((2n+1)(2n+2))) ⇒ Π_(k=0) ^(n−1) (a_(k+1) /a_k ) =Π_(k=0) ^(n−1) (1/((2k+1)(2k+2))) ⇒ (a_1 /a_0 )×(a_2 /a_1 )×....×(a_n /a_(n−1) )=(1/(Π_(k=0) ^(n−1) (2k+1)Π_(k=0) ^(n−1) (2k+2))) ⇒ a_n =a_0 ×(1/(Π_(k=0) ^(n−1) (2k+1)Π_(k=0) ^(n−1) (2k+2))) but Π_(k=0) ^(n−1) (2k+1)=1.3.5...(2n−1) =1.2.3.4.5.....(2n−1).2n×(1/(2.4...(2n))) =(((2n)!)/(2^n n!)) also Π_(k=0) ^(n−1) (2k+2) =Π_(k=1) ^n (2k) =2^n n! ⇒ a_n =a_0 .((2^n n!)/((2n)!)).(1/(2^n n!)) ⇒a_n =(a_o /((2n)!))
$$\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)}\:\Rightarrow \\ $$$$\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{{a}_{{k}+\mathrm{1}} }{{a}_{{k}} }\:=\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{2}\right)}\:\Rightarrow \\ $$$$\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{0}} }×\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }×….×\frac{{a}_{{n}} }{{a}_{{n}−\mathrm{1}} }=\frac{\mathrm{1}}{\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{1}\right)\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{2}\right)} \\ $$$$\Rightarrow\:{a}_{{n}} ={a}_{\mathrm{0}} ×\frac{\mathrm{1}}{\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{1}\right)\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{2}\right)} \\ $$$${but}\:\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{1}\right)=\mathrm{1}.\mathrm{3}.\mathrm{5}…\left(\mathrm{2}{n}−\mathrm{1}\right) \\ $$$$=\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}…..\left(\mathrm{2}{n}−\mathrm{1}\right).\mathrm{2}{n}×\frac{\mathrm{1}}{\mathrm{2}.\mathrm{4}…\left(\mathrm{2}{n}\right)} \\ $$$$=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} {n}!}\:{also} \\ $$$$\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{2}\right)\:=\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{2}{k}\right) \\ $$$$=\mathrm{2}^{{n}} {n}!\:\Rightarrow \\ $$$${a}_{{n}} ={a}_{\mathrm{0}} .\frac{\mathrm{2}^{{n}} {n}!}{\left(\mathrm{2}{n}\right)!}.\frac{\mathrm{1}}{\mathrm{2}^{{n}} {n}!}\:\Rightarrow{a}_{{n}} =\frac{{a}_{{o}} }{\left(\mathrm{2}{n}\right)!} \\ $$

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