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()

王守恩

(*定义数列：Fibonacci[k+2] 给出 2,3,5,8,13,...*)A[k_] := Fibonacci[k + 2]

(*生成第 n 行的数字列表*)GenerateList[1] := {1}
GenerateList[2] := {2, 1}
GenerateList[n_] := GenerateList[n] = Module[{prev, k}, If[EvenQ[n], prev = GenerateList[n - 1];
    Append[prev, Last[prev]], prev = GenerateList[n - 1]; prev = Most[prev]; k = (n - 1)/2; Join[prev, {A[k + 1], A[k]}]]]

(*构建乘积部分的字符串：返回类似 "Sqrt[2] * Sqrt[5 - 3]" 或 "Sqrt[2] * Sqrt[5 - \Sqrt[3] * Sqrt[3]]" 等*)
buildProductString[L_List] :=
Module[{x = L[[1]], y = L[[2]], rest = L[[3 ;;]]},
  Which[Length[rest] == 0,
   "Sqrt[" <> ToString[x] <> "] * Sqrt[" <> ToString[y] <> "]",
   Length[rest] == 1,
   "Sqrt[" <> ToString[x] <> "] * Sqrt[" <> ToString[y] <> " - " <>
    ToString[rest[[1]]] <> "]", True,
   "Sqrt[" <> ToString[x] <> "] * Sqrt[" <> ToString[y] <> " - " <>
    buildProductString[rest] <> "]"]]

(*将数字列表转换为字符串表达式，然后转为带 HoldForm 的排版表达式*)
MakeExpr[L_List] := Module[{str},
  str = Switch[Length[L], 1, ToString[L[[1]]], 2,
    "Sqrt[" <> ToString[L[[1]]] <> " - " <> ToString[L[[2]]] <> "]",
    3, "Sqrt[" <> ToString[L[[1]]] <> " - Sqrt[" <> ToString[L[[2]]] <>
      " - " <> ToString[L[[3]]] <> "]]", _,
    Module[{a = L[[1]], b = L[[2]], c = L[[3]], rest = L[[4 ;;]]},
     "Sqrt[" <> ToString[a] <> " - Sqrt[" <> ToString <> " - " <>
      buildProductString[Prepend[rest, c]] <> "]]"]];
  (*关键：使用 ToExpression 并强制用 HoldForm 包裹，防止任何化简*)ToExpression[str, StandardForm, HoldForm]]

(*显示第 n 行等式*)ShowG[n_] := Row[{"G(", n, ") = ", MakeExpr[GenerateList[n]], " = 1"}]

(*竖向排列显示 G(1) 到 G(18)*)Column[Table[ShowG, {i, 1, 18}], Alignment -> Left]