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Question Number 98673 by MJS last updated on 15/Jun/20
find a_n in terms of n (I can′t find it...) a_1 =1; a_2 =4 a_3 =a_2 ×4×((2^2 −1)/2^2 ) a_4 =a_3 ×4×((2^2 −1)/2^2 )×((3^2 −1)/3^2 ) a_5 =a_4 ×4×((2^2 −1)/2^2 )×((3^2 −1)/3^2 )×((4^2 −1)/4^2 ) ... n≥2: a_(n+1) =4a_n Π_(k=2) ^n ((k^2 −1)/k^2 )
$$\mathrm{find}\:{a}_{{n}} \:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n} \\ $$$$\left(\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\mathrm{it}…\right) \\ $$$${a}_{\mathrm{1}} =\mathrm{1};\:{a}_{\mathrm{2}} =\mathrm{4} \\ $$$${a}_{\mathrm{3}} ={a}_{\mathrm{2}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{4}} ={a}_{\mathrm{3}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{5}} ={a}_{\mathrm{4}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }×\frac{\mathrm{4}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}^{\mathrm{2}} } \\ $$$$… \\ $$$${n}\geqslant\mathrm{2}:\:{a}_{{n}+\mathrm{1}} =\mathrm{4}{a}_{{n}} \underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$
Answered by  M±th+et+s last updated on 16/Jun/20
sir what do you think about it Σ_(k=2) ^n ((k^2 −1)/k^2 )=(1/(2+((a_n −1)/(an+1))))
$${sir}\:{what}\:{do}\:{you}\:{think}\:{about}\:{it} \\ $$$$\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}+\frac{{a}_{{n}} −\mathrm{1}}{{an}+\mathrm{1}}} \\ $$
Answered by maths mind last updated on 16/Jun/20
⇔(a_(n+1) /a_n )=4.Π_(k=2) ^n .(((k−1)(k+1))/(k.k)). Π_(k=2) ^n ((k+1)/k)=((n+1)/2) Π_(k=2) ^n ((k−1)/k)=(1/n) ⇒(a_(n+1) /a_n )=((2(n+1))/n) ⇒Π_(j=1) ^(m−1) (a_(j+1) /a_j )=Π_(j=1) ^(m−1) ((2(n+1))/n)⇒(a_m /a_1 )=2^(m−1) .m ⇒a_m =a_1 .m.2^(m−1)
$$\Leftrightarrow\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\mathrm{4}.\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}.\frac{\left({k}−\mathrm{1}\right)\left({k}+\mathrm{1}\right)}{{k}.{k}}. \\ $$$$\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\frac{{k}+\mathrm{1}}{{k}}=\frac{{n}+\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\frac{{k}−\mathrm{1}}{{k}}=\frac{\mathrm{1}}{{n}} \\ $$$$\Rightarrow\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{{n}} \\ $$$$\Rightarrow\underset{{j}=\mathrm{1}} {\overset{{m}−\mathrm{1}} {\prod}}\frac{{a}_{{j}+\mathrm{1}} }{{a}_{{j}} }=\underset{{j}=\mathrm{1}} {\overset{{m}−\mathrm{1}} {\prod}}\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{{n}}\Rightarrow\frac{{a}_{{m}} }{{a}_{\mathrm{1}} }=\mathrm{2}^{{m}−\mathrm{1}} .{m} \\ $$$$\Rightarrow{a}_{{m}} ={a}_{\mathrm{1}} .{m}.\mathrm{2}^{{m}−\mathrm{1}} \\ $$
Commented by MJS last updated on 16/Jun/20
great, thank you! I hadn′t thought of this
$$\mathrm{great},\:\mathrm{thank}\:\mathrm{you}!\:\mathrm{I}\:\mathrm{hadn}'\mathrm{t}\:\mathrm{thought}\:\mathrm{of}\:\mathrm{this} \\ $$

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