\u4eca\u56de\u306f\u3053\u3061\u3089\u306e\u554f\u984c\u3092\u89e3\u3044\u3066\u3044\u304d\u307e\u3059.<\/p>\n\n\n\n
\u6570\u5217\\(\\left\\{a_n\\right\\}\\)\u3092 \u6570\u5217\\(\\left\\{a_n\\right\\}\\)\u306f\u30d5\u30a3\u30dc\u30ca\u30c3\u30c1\u6570\u5217\u3068\u547c\u3070\u308c\u308b\u6709\u540d\u306a\u6570\u5217\u3067\u3059. \u3053\u306e\u554f\u984c\u306f\u30d5\u30a3\u30dc\u30ca\u30c3\u30c1\u6570\u5217\u304c\\(\\tan\\)\u3092\u901a\u3058\u3066\u9762\u767d\u3044\u6027\u8cea\u304c\u3042\u308b\u3053\u3068\u3092\u793a\u5506\u3057\u3066\u3044\u307e\u3059. (1) \\(c_n=a_{n+1}a_{n-1}-a_n^2\\) \\((n\\geq 2)\\)\u3068\u304a\u304f\u3068, (2) \\(\\tan\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u3092\u7528\u3044\u3066, (3) (2)\u3088\u308a, \u6b63\u306e\u6574\u6570\\(m\\)\u306b\u5bfe\u3057\u3066, \u3053\u3061\u3089\u306e(1)\u306e\u89e3\u7b54\u3067\u306f\\(c_n\\)\u304c\u7b49\u6bd4\u6570\u5217\u3067\u3042\u308b\u3053\u3068\u3092\u898b\u629c\u304d\u307e\u3057\u305f\u304c, \u305d\u306e\u5909\u5f62\u3082\u5c11\u3057\u6280\u5de7\u7684\u3067\u81ea\u5206\u3067\u601d\u3044\u3064\u304f\u306e\u306f\u591a\u5c11\u96e3\u3057\u3044\u304b\u3068\u601d\u3044\u307e\u3059. \u3068\u3044\u3046\u3053\u3068\u3067\u4e88\u544a\u901a\u308a(1)\u306e\u5225\u89e3\u3092\u7d39\u4ecb\u3057\u307e\u3059. \u3053\u308c\u304b\u3089\u7d39\u4ecb\u3059\u308b\u65b9\u6cd5\u306f, \u300c\\(n\\)\u304c\u5c0f\u3055\u3044\u7bc4\u56f2\u3067\u5b9f\u969b\u306b\u5024\u3092\u8a08\u7b97\u3057, \u5f0f\u306e\u5f62\u3092\u4e88\u60f3\u3057, \u305d\u306e\u4e88\u60f3\u304c\u6b63\u3057\u3044\u3053\u3068\u3092\u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3067\u8a3c\u660e\u3059\u308b\u300d\u89e3\u6cd5\u3067\u3059. \u5177\u4f53\u7684\u306b\u6570\u5024\u3092\u8a08\u7b97\u3059\u308b\u70b9\u3084, \u5e30\u7d0d\u6cd5\u306e\u4eee\u5b9a\u3067\u4f7f\u3048\u308b\u6570\u5f0f\u304c\u6c7a\u307e\u3063\u3066\u3044\u308b\u5206, \u4e0a\u306e\u89e3\u7b54\u3088\u308a\u304b\u306f\u8fbf\u308a\u3064\u304d\u3084\u3059\u3044\u89e3\u7b54\u3067\u306f\u306a\u3044\u304b\u3068\u601d\u3044\u307e\u3059.<\/p>\n\n\n\n (1)\u306e\u5225\u89e3) youtube\u3067\u3082\u89e3\u8aac\u3057\u3066\u3044\u307e\u3059.<\/p>\n\n\n\n
$$
\\begin{align}
a_1&=a_2=1\\\\[1.5ex]
a_{n+2}&=a_{n+1}+a_n\\,\\,\\,(n=1,2,3,\\cdots)
\\end{align}
$$
\u306b\u3088\u308a\u5b9a\u3081\u308b. \u6b21\u306b, \u6570\u5217\\(\\left\\{b_n\\right\\}\\)\u3092
$$
\\tan{b_n}=\\frac{1}{a_n}
$$\u306b\u3088\u308a\u5b9a\u3081\u308b. \u305f\u3060\u3057, \\( \\displaystyle 0<b_n<\\frac{\\pi}{2}\\)\u3068\u3059\u308b.
\u3053\u306e\u3068\u304d, \u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.
(1) \\(n\\geq 2\\)\u306b\u5bfe\u3057\u3066, \\(\\displaystyle a_{n+1}a_{n-1}-a_n^2\\)\u3092\u6c42\u3081\u3088.
(2) \u6b63\u306e\u6574\u6570\\(m\\)\u306b\u5bfe\u3057\u3066, \\( \\displaystyle a_{2m}\\cdot\\tan{\\left(b_{2m+1}+b_{2m+2}\\right)}\\)\u3092\u6c42\u3081\u3088.
(3) \\(\\displaystyle \\sum_{m=0}^\\infty b_{2m+1}\\)\u306e\u5024\u3092\u6c42\u3081\u3088.
(2025 \u6771\u4eac\u79d1\u5b66\u5927\u5b66 [4])<\/span><\/p>\n\n\n\n
\u306a\u304a, \u4ee5\u4e0b\u306e\u89e3\u7b54\u306e(1)\u306f\u601d\u3044\u3064\u304d\u306b\u304f\u3044\u304b\u3082\u3057\u308c\u307e\u305b\u3093. \u305d\u3053\u3067, \u300c\\(n\\)\u304c\u5c0f\u3055\u3044\u7bc4\u56f2\u3067\u5b9f\u969b\u306b\u5024\u3092\u8a08\u7b97\u3057, \u5f0f\u306e\u5f62\u3092\u4e88\u60f3\u3057, \u305d\u306e\u4e88\u60f3\u304c\u6b63\u3057\u3044\u3053\u3068\u3092\u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3067\u8a3c\u660e\u3059\u308b\u300d\u3068\u3044\u3046\u5225\u89e3\u3092\u89e3\u7b54\u306e\u5f8c\u306b\u8f09\u305b\u3066\u304a\u308a\u307e\u3059\u306e\u3067, \u305d\u3061\u3089\u3082\u5408\u308f\u305b\u3066\u3054\u53c2\u7167\u304f\u3060\u3055\u3044.
\u305d\u308c\u3067\u306f\u89e3\u3044\u3066\u3044\u304d\u307e\u3057\u3087\u3046.<\/p>\n\n\n\n
$$
\\begin{align}
c_{n+1}&=a_{n+1}a_{n-1}-a_n^2=(a_{n}+a_{n-1})a_{n-1}-a_n(a_{n-1}+a_{n-2})\\\\[1.5ex]
&=-(a_na_{n-2}-a_{n-1}^2)=-c_n
\\end{align}
$$\u3068\u306a\u308b. \u3088\u3063\u3066, \\(\\left\\{c_n\\right\\} \\) \\((n\\geq 2)\\)\u306f\u516c\u6bd4\\(-1\\)\u306e\u7b49\u6bd4\u6570\u5217\u3067\u3042\u308b. \u3053\u308c\u304b\u3089,
$$
c_n=-c_{n-1}=(-1)^2c_{n-2}=\\cdots=(-1)^{n-2}c_2=(-1)^nc_2
$$\u3068\u306a\u308b. \u3053\u3053\u3067
$$
c_2=a_3a_1-a_2^2=(a_2+a_1)a_1-a_2^2=(1+1)\\cdot 1-1^2=1
$$\u3088\u308a, \\(\\displaystyle c_n=(-1)^n\\)\u3068\u306a\u308a,
$$
a_{n+1}a_{n-1}-a_n^2=(-1)^n,\\,\\,(n\\geq 2)
$$\u304c\u308f\u304b\u308b.<\/p>\n\n\n\n
$$
\\begin{align}
a_{2m}\\cdot\\tan{\\left(b_{2m+1}+b_{2m+2}\\right)}
&=a_{2m}\\cdot\\frac{\\tan{b_{2m+1}}+\\tan{b_{2m+2}}}{1-\\tan{b_{2m+1}}\\tan{b_{2m+2}}} \\\\[1.5ex]
&=a_{2m}\\cdot\\frac{\\frac{1}{a_{2m+1}}+\\frac{1}{a_{2m+2}}}{1-\\frac{1}{a_{2m+1}}\\cdot\\frac{1}{a_{2m+2}}}\\\\[1.5ex]
&=a_{2m}\\cdot\\frac{a_{2m+2}+a_{2m+1}}{a_{2m+1}a_{2m+2}-1}\\\\[1.5ex]
&=\\frac{a_{2m}a_{2m+2}+a_{2m}a_{2m+1}}{a_{2m+1}a_{2m+2}-1}\\\\[1.5ex]
&=\\frac{a_{2m+1}^2+(-1)^{2m+1}+a_{2m}a_{2m+1}}{a_{2m+1}a_{2m+2}-1}\\\\[1.5ex]
&=\\frac{a_{2m+1}(a_{2m+1}+a_{2m})-1}{a_{2m+1}a_{2m+2}-1}\\\\[1.5ex]
&=\\frac{a_{2m+1}a_{2m+2}-1}{a_{2m+1}a_{2m+2}-1}=1
\\end{align}
$$\u3068\u3057\u3066\u6c42\u307e\u308b. \u305f\u3060\u3057\u9014\u4e2d\u3067, (1)\u304b\u3089\u308f\u304b\u308b,
$$
a_{2m}a_{2m+2}=a_{2m+1}^2+(-1)^{2m+1}
$$\u3092\u7528\u3044\u305f.<\/p>\n\n\n\n
$$
\\tan{\\left(b_{2m+1}+b_{2m+2}\\right)}=\\frac{1}{a_{2m}}=\\tan{b_{2m}}
$$\u3068\u306a\u308b.
\u3053\u3053\u3067, \\(\\left\\{b_n\\right\\}\\)\u306e\u5b9a\u7fa9\u304b\u3089, \\(\\displaystyle 0<b_{2m}<\\frac{\\pi}{2}\\)\u3067\u3042\u308b\u304b\u3089,
$$
\\tan{\\left(b_{2m+1}+b_{2m+2}\\right)}=\\tan{b_{2m}}>0
$$\u304c\u308f\u304b\u308b. \u3055\u3089\u306b, \\(0<b_{2m+1}+b_{2m+2}<\\pi\\)\u3067\u3042\u308a, \\(\\tan{\\left(b_{2m+1}+b_{2m+2}\\right)}>0\\)\u3067\u3042\u308b\u3053\u3068\u304b\u3089, $$
0<b_{2m+1}+b_{2m+2}<\\frac{\\pi}{2}
$$\u304c\u308f\u304b\u308b.
\\(\\displaystyle 0<x<\\frac{\\pi}{2}\\)\u306e\u7bc4\u56f2\u3067\u306f\\(\\tan{x}\\)\u306f\u5358\u8abf\u5897\u52a0\u3067\u3042\u308b\u304b\u3089,
$$
\\tan{\\left(b_{2m+1}+b_{2m+2}\\right)}=\\tan{b_{2m}} \\Longrightarrow b_{2m+1}+b_{2m+2}=b_{2m}
$$\u304c\u308f\u304b\u308b. \u3088\u3063\u3066\u3053\u308c\u304b\u3089,
$$
b_{2m+1}=b_{2m}-b_{2m+2},\\,\\,(m\\geq 1)
$$\u3068\u306a\u308b.
\u307e\u305a\u306f\u6c42\u3081\u308b\u7121\u9650\u7d1a\u6570\u306e\\(m=N\\)\u307e\u3067\u306e\u6709\u9650\u548c\u3092\u8003\u3048\u308b\u3068,
$$
\\begin{align}
\\sum_{m=0}^Nb_{2m+1}&=b_1+\\sum_{m=0}^Nb_{2m+1}=b_1+\\sum_{m=0}^N\\left(b_{2m}-b_{2m+2}\\right)\\\\[1.5ex]
&=b_1+(b_2-b_4)+(b_4-b_6)+\\cdots +(b_{2N}-b_{2N+2})\\\\[1.5ex]
&=b_1+(b_2-\\cancel{b_4})+(\\cancel{b_4}-\\cancel{b_6})+\\cdots +(\\cancel{b_{2N}}-b_{2N+2})\\\\[1.5ex]
&=b_1+b_2-b_{2N+2}
\\end{align}
$$\u3068\u306a\u308b.
\u307e\u305a, \\(b_1\\), \\(b_2\\)\u306f
$$
\\tan{b_1}=\\frac{1}{a_1}=1,\\,\\,\\tan{b_2}=\\frac{1}{a_2}=1\\\\
$$\u3088\u308a,
$$
b_1=b_2=\\frac{\\pi}{4}
$$\u304c\u308f\u304b\u308b.
\u6b21\u306b\\(b_{2N+2}\\)\u306e\\(N\\rightarrow \\infty\\)\u306e\u3068\u304d\u306e\u6975\u9650\u3092\u77e5\u308a\u305f\u3044\u306e\u3067, \\(b_n\\)\u306e\u6975\u9650\u3092\u6c42\u3081\u308b. \\(\\left\\{a_n\\right\\}\\)\u306f\u305d\u306e\u4f5c\u308a\u65b9\u304b\u3089\\(a_n\\geq1\\)\u3092\u6e80\u305f\u3059\u306e\u3067,
$$
\\begin{align}
a_n&=a_{n-1}+a_{n-2}\\geq a_{n-1}+1=a_{n-2}+a_{n-1}+1\\\\[1.5ex]
&\\geq a_{n-2}+2\\geq \\cdots \\geq a_1+(n-1)=n
\\end{align}
$$\u3068\u306a\u308b\u304b\u3089, $$
0<\\tan{b_n}=\\frac{1}{a_n}\\leq\\frac{1}{n}
$$\u3067\u3042\u308b. \u3088\u3063\u3066, \u306f\u3055\u307f\u3046\u3061\u306e\u539f\u7406\u304b\u3089
$$
\\lim_{n\\rightarrow \\infty}\\tan{b_n}=0
$$\u304c\u308f\u304b\u308a, \u3053\u308c\u304b\u3089,
$$
\\lim_{n\\rightarrow \\infty}{b_n}=0
$$\u3068\u306a\u308b.
\u4ee5\u4e0a\u304b\u3089,
$$
\\begin{align}
\\sum_{m=0}^\\infty b_{2m+1}&=\\lim_{N\\rightarrow \\infty}\\sum_{m=0}^N b_{2m+1}\\\\[1.5ex]
&=\\lim_{N\\rightarrow \\infty}\\left(b_1+b_2+b_{2N+2}\\right)\\\\[1.5ex]
&=b_1+b_2+\\lim_{N\\rightarrow \\infty}b_{2N+2}=\\frac{\\pi}{4}+\\frac{\\pi}{4}+0=\\frac{\\pi}{2}
\\end{align}
$$\u3068\u6c42\u307e\u308b.<\/p>\n<\/div><\/div>\n\n\n\n
\\(a_1=1\\), \\(a_2=1\\), \\(a_3=2\\), \\(a_4=3\\), \\(a_5=5\\), \\(a_6=8\\)\u306b\u6ce8\u610f\u3057\u3066, \\(n=2, 3, 4, 5, \\)\u306b\u5bfe\u3057\u3066, \\(a_{n+1}a_{n-1}-a_n^2\\)\u3092\u8a08\u7b97\u3057\u3066\u3044\u304f\u3068,
$$
\\begin{align}
a_{3}a_{1}-a_2^2&=2\\cdot 1-1^2=1\\\\[1.5ex]
a_{4}a_{2}-a_3^2&=3\\cdot 1-2^2=-1\\\\[1.5ex]
a_{5}a_{3}-a_4^2&=5\\cdot 2-3^2=1\\\\[1.5ex]
a_{6}a_{4}-a_5^2&=8\\cdot 3-5^2=-1
\\end{align}
$$\u3067\u3042\u308b\u304b\u3089,
$$
a_{n+1}a_{n-1}-a_n^2=(-1)^n,\\,\\,(n=2,3,4,\\cdots) \u30fb\u30fb\u30fb\u2460
$$\u3067\u3042\u308b\u3053\u3068\u304c\u4e88\u60f3\u3067\u304d\u308b. \u3053\u308c\u3092\u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3067\u8a3c\u660e\u3059\u308b.
1) \\(n=2\\)\u306e\u3068\u304d\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u3059
$$
\\begin{align}
(\u5de6\u8fba)&=a_{3}a_{1}-a_2^2=2\\cdot 1-1^2=1\\\\[1.5ex]
(\u53f3\u8fba)&=(-1)^2=1
\\end{align}
$$\u3068\u306a\u308a\u6210\u308a\u7acb\u3064.
2) \\(n=k\\)\u306e\u3068\u304d\u306b\u6210\u308a\u7acb\u3064\u3068\u4eee\u5b9a\u3057\u3066, \\(n=k+1\\)\u306e\u3068\u304d\u306b\u3082\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u3059
\u307e\u305a, \u2460\u304c\\(n=k\\)\u306e\u3068\u304d\u6210\u308a\u7acb\u3064\u3068\u4eee\u5b9a\u3059\u308b. \u3064\u307e\u308a, \u4ee5\u4e0b\u306e\u7b49\u5f0f\u304c\u6210\u308a\u7acb\u3064\u3068\u3059\u308b.
$$
a_{k+1}a_{k-1}-a_k^2=(-1)^k
$$
\u3053\u306e\u524d\u63d0\u306e\u3082\u3068, \u2460\u3067\\(n=k+1\\)\u3068\u3057\u305f\u7b49\u5f0f\u306e\u5de6\u8fba\u306f
$$
\\begin{align}
a_{k+2}a_{k}-a_{k+1}^2&=\\left(a_{k+1}+a_{k}\\right)a_{k}-a_{k+1}^2\\\\[1.5ex]
&=a_{k+1}a_{k}+a_{k}^2-a_{k+1}^2
\\end{align}
$$ \u3068\u306a\u308b. \u3053\u3053\u3067, \u5e30\u7d0d\u6cd5\u306e\u4eee\u5b9a\u3088\u308a,
$$
a_k^2=a_{k+1}a_{k-1}-(-1)^k
$$\u3067\u3042\u308b\u304b\u3089, \u3053\u308c\u3092\u4ee3\u5165\u3057\u3066\u5f0f\u5909\u5f62\u3092\u7d9a\u3051\u308b\u3068
$$
\\begin{align}
a_{k+2}a_{k}-a_{k+1}^2&=a_{k+1}a_{k}+a_{k+1}a_{k-1}-(-1)^k-a_{k+1}^2\\\\[1.5ex]
&=a_{k+1}\\left(a_k+a_{k-1}\\right)-(-1)^k-a_{k+1}^2\\\\[1.5ex]
&=a_{k+1}^2-(-1)^k-a_{k+1}^2\\\\[1.5ex]
&=-(-1)^k=(-1)^{k+1}
\\end{align}
$$\u3068\u306a\u308a, \u3053\u308c\u306f\u2460\u3067\\(n=k+1\\)\u3068\u3057\u305f\u53f3\u8fba\u306b\u4e00\u81f4\u3059\u308b. \u3088\u3063\u3066\u2460\u306f\\(n=k+1\\)\u306e\u3068\u304d\u3082\u6210\u308a\u7acb\u3064.
1), 2)\u3088\u308a\u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u304b\u3089\\(n\\geq 2\\)\u3068\u306a\u308b\u81ea\u7136\u6570\\(n\\)\u306b\u5bfe\u3057\u3066,
$$
a_{n+1}a_{n-1}-a_n^2=(-1)^n
$$\u304c\u6210\u308a\u7acb\u3064.<\/p>\n\n\n\n