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3-1-6-3-2-6-5-2-6-7-2-6-9-2-6-provet-that-




Question Number 184969 by SEKRET last updated on 14/Jan/23
                3 + (1/(6+(3^2 /(6+(5^2 /(6+(7^2 /(6+(9^2 /(6+......))   ))  ))  ))    ))  = 𝛑      provet  that.
$$\:\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{3}\:+\:\frac{\mathrm{1}}{\mathrm{6}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{6}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{6}+\frac{\mathrm{7}^{\mathrm{2}} }{\mathrm{6}+\frac{\mathrm{9}^{\mathrm{2}} }{\mathrm{6}+……}\:\:\:}\:\:}\:\:}\:\:\:\:}\:\:=\:\boldsymbol{\pi} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{provet}}\:\:\boldsymbol{\mathrm{that}}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Frix last updated on 15/Jan/23
I know the proof for this one  π=(4/(1+(1^2 /(2+(3^2 /(2+(5^2 /(2+(7^2 /(2+...))))))))))  which was found by John Wallis in 1665.
$$\mathrm{I}\:\mathrm{know}\:\mathrm{the}\:\mathrm{proof}\:\mathrm{for}\:\mathrm{this}\:\mathrm{one} \\ $$$$\pi=\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{2}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{2}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{2}+\frac{\mathrm{7}^{\mathrm{2}} }{\mathrm{2}+…}}}}} \\ $$$$\mathrm{which}\:\mathrm{was}\:\mathrm{found}\:\mathrm{by}\:{John}\:{Wallis}\:\mathrm{in}\:\mathrm{1665}. \\ $$
Commented by SEKRET last updated on 15/Jan/23
??
$$?? \\ $$
Answered by aba last updated on 14/Jan/23
[3]=3  [3;6]=3+(1/6)=((19)/6)=3,166...  [3;6;6]=3+(1/(6+(9/6)))=((141)/(45))=3,1333...  [3;6;6;6]=3+(1/(6+(9/(6+((25)/6)))))=((1321)/(420))=3,145...  [3;6;6;6;6]=3+(1/(6+(9/(6+((25)/(6+((49)/6)))))))=((989)/(315))=3,13968...  [3;6;6;6;6;6]=3+(1/(6+(9/(6+((25)/(6+((49)/(6+((121)/6)))))))))=((248851)/(79170))=3,143...  [3;6;6;6;6..........∞]=π
$$\left[\mathrm{3}\right]=\mathrm{3} \\ $$$$\left[\mathrm{3};\mathrm{6}\right]=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{19}}{\mathrm{6}}=\mathrm{3},\mathrm{166}… \\ $$$$\left[\mathrm{3};\mathrm{6};\mathrm{6}\right]=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}+\frac{\mathrm{9}}{\mathrm{6}}}=\frac{\mathrm{141}}{\mathrm{45}}=\mathrm{3},\mathrm{1333}… \\ $$$$\left[\mathrm{3};\mathrm{6};\mathrm{6};\mathrm{6}\right]=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}+\frac{\mathrm{9}}{\mathrm{6}+\frac{\mathrm{25}}{\mathrm{6}}}}=\frac{\mathrm{1321}}{\mathrm{420}}=\mathrm{3},\mathrm{145}… \\ $$$$\left[\mathrm{3};\mathrm{6};\mathrm{6};\mathrm{6};\mathrm{6}\right]=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}+\frac{\mathrm{9}}{\mathrm{6}+\frac{\mathrm{25}}{\mathrm{6}+\frac{\mathrm{49}}{\mathrm{6}}}}}=\frac{\mathrm{989}}{\mathrm{315}}=\mathrm{3},\mathrm{13968}… \\ $$$$\left[\mathrm{3};\mathrm{6};\mathrm{6};\mathrm{6};\mathrm{6};\mathrm{6}\right]=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}+\frac{\mathrm{9}}{\mathrm{6}+\frac{\mathrm{25}}{\mathrm{6}+\frac{\mathrm{49}}{\mathrm{6}+\frac{\mathrm{121}}{\mathrm{6}}}}}}=\frac{\mathrm{248851}}{\mathrm{79170}}=\mathrm{3},\mathrm{143}… \\ $$$$\left[\mathrm{3};\mathrm{6};\mathrm{6};\mathrm{6};\mathrm{6}……….\infty\right]=\pi \\ $$
Commented by Frix last updated on 15/Jan/23
This is not a proof.
$$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{proof}. \\ $$
Commented by aba last updated on 15/Jan/23
this is a proposal
$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{proposal} \\ $$

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