兔子数列求和

王守恩

\[1=\sqrt{2-\sqrt{3-\sqrt{2}\sqrt{5-\sqrt{3}\sqrt{8-\sqrt{5}\sqrt{13-\sqrt{8}\sqrt{21-\sqrt{13}\sqrt{34-\sqrt{21}}}}}}}}\]
\[1=\sqrt{2-\sqrt{3-\sqrt{2}\sqrt{5-\sqrt{3}\sqrt{8-\sqrt{5}\sqrt{13-\sqrt{8}\sqrt{21-\sqrt{13}\sqrt{34-\sqrt{21}\sqrt{55-}}}}}}}}\]
\[1=\sqrt{2-\sqrt{3-\sqrt{2}\sqrt{5-\sqrt{3}\sqrt{8-\sqrt{5}\sqrt{13-\sqrt{8}\sqrt{21-\sqrt{13}\sqrt{34-\sqrt{21}\sqrt{55-\sqrt{34}}}}}}}}}\]
\[1=\sqrt{2-\sqrt{3-\sqrt{2}\sqrt{5-\sqrt{3}\sqrt{8-\sqrt{5}\sqrt{13-\sqrt{8}\sqrt{21-\sqrt{13}\sqrt{34-\sqrt{21}\sqrt{55-\sqrt{34}\sqrt{89-}}}}}}}}}\]
这是4道题(看最后1个数)，每道题(同时加4个数)的和都在向 “1” 靠拢，求助：
1，这算式太长了，在 mathematica 上应如何编写这算式？
2，在 mathematica 上，应如何编写，答案 “1” 才能出来？

aimisiyou

意义不大，根据递推关系式就可以得出结论。

王守恩

aimisiyou 发表于 2020-4-29 07:37
意义不大，根据递推关系式就可以得出结论。

请您也来一串？
1=  1
1=s(2 - 1]
1=s(2-s(3 - 2]
1=s(2-s(3-s(2)s(2]
1=s(2-s(3-s(2)s(5 - 3]
1=s(2-s(3-s(2)s(5-s(3)s(3]
1=s(2-s(3-s(2)s(5-s(3)s(8 - 5]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(5]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13 - 8]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(8]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21 - 13]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(13]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34 - 21]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(21]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55 - 34]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55-s(34)s(34]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55-s(34)s(89 - 55]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55-s(34)s(89-s(55)s(55]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55-s(34)s(89-s(55)s(144 - 89]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55-s(34)s(89-s(55)s(144-s(89)s(89]
s=sqrt，]=多重()