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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

$$\mathrm{The}\mathrm{sequence}\mathrm{seems}\mathrm{to}\mathrm{be}\mathrm{a}\mathrm{sequence}$$$$\mathrm{of}\mathrm{rational}\mathrm{numbers}\mathrm{and}\mathrm{the}\mathrm{limit}$$$$\mathrm{seems}\mathrm{to}\mathrm{tend}\mathrm{to}0.$$$$\mathrm{We}\mathrm{try}\mathrm{a}\mathrm{sequence}\mathrm{like}:$$$$u_n=\frac{an+b}{{cn}^2+dn+e}$$$$\mathrm{By}\mathrm{using}\mathrm{the}\mathrm{datas}{u}_0=-6,u_1=\frac{9}{2},$$$$u_3=\frac{3}{4},u_5=\frac{7}{18},\mathrm{we}\mathrm{find}:$$$$u_n=\frac{3n+6}{2n^2+n-1}$$$$\mathrm{We}\mathrm{can}\mathrm{verify}\mathrm{that}:$$$$u_0=\frac{3\times 0+6}{2\times 0^2+0-1}=-6$$$$u_1=\frac{3\times 1+6}{2\times 1^2+1-1}=\frac{9}{2}$$$$u_3=\frac{3\times 3+6}{2\times 3^2+3-1}=\frac{3}{4}$$$$u_5=\frac{3\times 5+6}{2\times 5^2+5-1}=\frac{7}{18}$$$$$$$$\bullet x=u_2=\frac{3\times 2+6}{2\times 2^2+2-1}=\frac{4}{3}$$$$\bullet y=u_4=\frac{3\times 4+6}{2\times 4^2+4-1}=\frac{18}{35}$$$$$$$$\bullet \mathrm{We}\mathrm{solve}{u}_n=\frac{3n+6}{2n^2+n-1}=\frac{8}{195}$$$$\mathrm{We}\mathrm{find}n=-\frac{31}{16}\mathrm{or}n=38$$$$\mathrm{Of}\mathrm{course}\mathrm{the}\mathrm{only}\mathrm{possible}\mathrm{solution}$$$$\mathrm{is}n=38.$$$$\Rightarrow \frac{8}{195}\mathrm{is}\mathrm{the}{39}^{\mathrm{th}}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{sequence}$$

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