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Question-148541

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Question Number 148541 by gloriousman last updated on 29/Jul/21
Answered by Olaf_Thorendsen last updated on 29/Jul/21
The sequence seems to be a sequence of rational numbers and the limit seems to tend to 0. We try a sequence like : u_n = ((an+b)/(cn^2 +dn+e)) By using the datas u_0 = −6, u_1 = (9/2), u_3 = (3/4), u_5 = (7/(18)), we find : u_n = ((3n+6)/(2n^2 +n−1)) We can verify that : u_0 = ((3×0+6)/(2×0^2 +0−1)) = −6 u_1 = ((3×1+6)/(2×1^2 +1−1)) = (9/2) u_3 = ((3×3+6)/(2×3^2 +3−1)) = (3/4) u_5 = ((3×5+6)/(2×5^2 +5−1)) = (7/(18)) • x = u_2 = ((3×2+6)/(2×2^2 +2−1)) = (4/3) • y = u_4 = ((3×4+6)/(2×4^2 +4−1)) = ((18)/(35)) • We solve u_n = ((3n+6)/(2n^2 +n−1)) = (8/(195)) We find n = −((31)/(16)) or n = 38 Of course the only possible solution is n = 38. ⇒ (8/(195)) is the 39^(th) term of the sequence
$$\mathrm{The}\:\mathrm{sequence}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a}\:\mathrm{sequence} \\ $$$$\mathrm{of}\:\mathrm{rational}\:\mathrm{numbers}\:\mathrm{and}\:\mathrm{the}\:\mathrm{limit} \\ $$$$\mathrm{seems}\:\mathrm{to}\:\mathrm{tend}\:\mathrm{to}\:\mathrm{0}. \\ $$$$\mathrm{We}\:\mathrm{try}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{like}\:: \\ $$$${u}_{{n}} \:=\:\frac{{an}+{b}}{{cn}^{\mathrm{2}} +{dn}+{e}} \\ $$$$\mathrm{By}\:\mathrm{using}\:\mathrm{the}\:\mathrm{datas}\:{u}_{\mathrm{0}} \:=\:−\mathrm{6},\:{u}_{\mathrm{1}} \:=\:\frac{\mathrm{9}}{\mathrm{2}}, \\ $$$${u}_{\mathrm{3}} \:=\:\frac{\mathrm{3}}{\mathrm{4}},\:{u}_{\mathrm{5}} \:=\:\frac{\mathrm{7}}{\mathrm{18}},\:\mathrm{we}\:\mathrm{find}\:: \\ $$$${u}_{{n}} \:=\:\frac{\mathrm{3}{n}+\mathrm{6}}{\mathrm{2}{n}^{\mathrm{2}} +{n}−\mathrm{1}} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{verify}\:\mathrm{that}\:: \\ $$$${u}_{\mathrm{0}} \:=\:\frac{\mathrm{3}×\mathrm{0}+\mathrm{6}}{\mathrm{2}×\mathrm{0}^{\mathrm{2}} +\mathrm{0}−\mathrm{1}}\:=\:−\mathrm{6} \\ $$$${u}_{\mathrm{1}} \:=\:\frac{\mathrm{3}×\mathrm{1}+\mathrm{6}}{\mathrm{2}×\mathrm{1}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}\:=\:\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${u}_{\mathrm{3}} \:=\:\frac{\mathrm{3}×\mathrm{3}+\mathrm{6}}{\mathrm{2}×\mathrm{3}^{\mathrm{2}} +\mathrm{3}−\mathrm{1}}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${u}_{\mathrm{5}} \:=\:\frac{\mathrm{3}×\mathrm{5}+\mathrm{6}}{\mathrm{2}×\mathrm{5}^{\mathrm{2}} +\mathrm{5}−\mathrm{1}}\:=\:\frac{\mathrm{7}}{\mathrm{18}} \\ $$$$ \\ $$$$\bullet\:{x}\:=\:{u}_{\mathrm{2}} \:=\:\frac{\mathrm{3}×\mathrm{2}+\mathrm{6}}{\mathrm{2}×\mathrm{2}^{\mathrm{2}} +\mathrm{2}−\mathrm{1}}\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\bullet\:{y}\:=\:{u}_{\mathrm{4}} \:=\:\frac{\mathrm{3}×\mathrm{4}+\mathrm{6}}{\mathrm{2}×\mathrm{4}^{\mathrm{2}} +\mathrm{4}−\mathrm{1}}\:=\:\frac{\mathrm{18}}{\mathrm{35}} \\ $$$$ \\ $$$$\bullet\:\mathrm{We}\:\mathrm{solve}\:{u}_{{n}} \:=\:\frac{\mathrm{3}{n}+\mathrm{6}}{\mathrm{2}{n}^{\mathrm{2}} +{n}−\mathrm{1}}\:=\:\frac{\mathrm{8}}{\mathrm{195}} \\ $$$$\mathrm{We}\:\mathrm{find}\:{n}\:=\:−\frac{\mathrm{31}}{\mathrm{16}}\:\mathrm{or}\:{n}\:=\:\mathrm{38} \\ $$$$\mathrm{Of}\:\mathrm{course}\:\mathrm{the}\:\mathrm{only}\:\mathrm{possible}\:\mathrm{solution} \\ $$$$\mathrm{is}\:{n}\:=\:\mathrm{38}. \\ $$$$\Rightarrow\:\frac{\mathrm{8}}{\mathrm{195}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{39}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sequence} \\ $$

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