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Question-211414




Question Number 211414 by AlagaIbile last updated on 08/Sep/24
Answered by Frix last updated on 08/Sep/24
Let x_j ≤x_(j+1)   (1/x_1 )=1 ⇒ max (x_1 ) =1  (1/x_1 )+(1/x_2 )=1 ⇒ max (x_2 ) =1×(1+1)=2  (1/x_1 )+(1/x_2 )+(1/x_3 )=1 ⇒ max (x_3 ) =2×(2+1)=6  ...  max (x_(k+1) ) =max (x_k ) ×(max (x_k ) +1)  x_6 =3263442
$$\mathrm{Let}\:{x}_{{j}} \leqslant{x}_{{j}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{{x}_{\mathrm{1}} }=\mathrm{1}\:\Rightarrow\:\mathrm{max}\:\left({x}_{\mathrm{1}} \right)\:=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{x}_{\mathrm{1}} }+\frac{\mathrm{1}}{{x}_{\mathrm{2}} }=\mathrm{1}\:\Rightarrow\:\mathrm{max}\:\left({x}_{\mathrm{2}} \right)\:=\mathrm{1}×\left(\mathrm{1}+\mathrm{1}\right)=\mathrm{2} \\ $$$$\frac{\mathrm{1}}{{x}_{\mathrm{1}} }+\frac{\mathrm{1}}{{x}_{\mathrm{2}} }+\frac{\mathrm{1}}{{x}_{\mathrm{3}} }=\mathrm{1}\:\Rightarrow\:\mathrm{max}\:\left({x}_{\mathrm{3}} \right)\:=\mathrm{2}×\left(\mathrm{2}+\mathrm{1}\right)=\mathrm{6} \\ $$$$… \\ $$$$\mathrm{max}\:\left({x}_{{k}+\mathrm{1}} \right)\:=\mathrm{max}\:\left({x}_{{k}} \right)\:×\left(\mathrm{max}\:\left({x}_{{k}} \right)\:+\mathrm{1}\right) \\ $$$${x}_{\mathrm{6}} =\mathrm{3263442} \\ $$
Commented by AlagaIbile last updated on 08/Sep/24
 Yes. Sylvester′s Sequence is the hint
$$\:{Yes}.\:{Sylvester}'{s}\:{Sequence}\:{is}\:{the}\:{hint} \\ $$

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