兔子数列求和
本帖最后由 王守恩 于 2020-4-29 13:47 编辑\
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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？ 意义不大，根据递推关系式就可以得出结论。 本帖最后由 王守恩 于 2020-4-30 15:22 编辑
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()
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