\u6570\u5217 \\(\\{ a _ n \\} , \\{ b _ n \\}\\) \u3092\u6b21\u306e\u3088\u3046\u306b\u5b9a\u7fa9\u3059\u308b.\r\n\\[\r\n\\left\\{ \\begin{array}{ll} a _ 1 =5 , \\ b _ 1 =3 , & \\\\ \\left( \\begin{array}{c} a _ {n+1} \\\\ b _ {n+1} \\end{array} \\right) = \\left( \\begin{array}{cc} 5 & 3 \\\\ 3 & 5 \\end{array} \\right) \\left( \\begin{array}{c} a _ n \\\\ b _ n \\end{array} \\right) & (n=1, 2, 3, \\cdots ) \\end{array} \\right.\r\n\\]\r\n\u307e\u305f, \u81ea\u7136\u6570 \\(n\\) \u306b\u3064\u3044\u3066 \\(c _ n ={a _ n}^2 -{b _ n}^2\\) \u3068\u304a\u304f. \u3053\u306e\u3068\u304d\u4ee5\u4e0b\u306e\u5404\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n
(1)<\/strong>\u3000\\(c _ n\\) \u3092 \\(n\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(k\\) \u3092\u81ea\u7136\u6570\u3068\u3059\u308b\u3068\u304d, \u81ea\u7136\u6570 \\(\\ell\\) \u306b\u3064\u3044\u3066\r\n\\[\r\na _ {k+\\ell} =a _ k a _ {\\ell} +b _ k b _ {\\ell} , \\ b _ {k+\\ell} =b _ k a _ {\\ell} +a _ k b _ {\\ell}\r\n\\]\r\n\u304c\u6210\u7acb\u3059\u308b\u3053\u3068\u3092, \\(\\ell\\) \u306b\u95a2\u3059\u308b\u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u306b\u3088\u3063\u3066\u793a\u305b.<\/p><\/li>\r\n (3)<\/strong>\u3000\\(n \\gt \\ell\\) \u3068\u306a\u308b\u81ea\u7136\u6570 \\(n , \\ell\\) \u306b\u3064\u3044\u3066\r\n\\[\r\nb _ {n+\\ell} -c _ {\\ell} b _ {n-\\ell} = 2a _ n b _ {\\ell}\r\n\\]\r\n\u304c\u6210\u7acb\u3059\u308b\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n (4)<\/strong>\u3000\\(2\\) \u4ee5\u4e0a\u306e\u81ea\u7136\u6570 \\(n\\) \u306b\u3064\u3044\u3066\r\n\\[\r\na _ {2n} +\\textstyle\\sum\\limits _ {m=1}^{n-1} c _ {n-m} a _ {2m} =\\dfrac{b _ {2n+1}}{2b _ 1} -\\dfrac{c _ n}{2}\r\n\\]\r\n\u304c\u6210\u7acb\u3059\u308b\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \u6761\u4ef6\u3088\u308a\r\n\\[\r\na _ {n+1} = 5a _ n +3b _ n , \\ b _ {n+1} = 3a _ n +5b _ n\n\\]\r\n\u306a\u306e\u3067\r\n\\[\\begin{align}\r\nc _ {n+1} & = {a _ {n+1}}^2 -{b _ {n+1}}^2 \\\\\r\n& = \\left( 5a _ n +3b _ n \\right)^2 -\\left( 3a _ n +5b _ n \\right)^2 \\\\\r\n& = 16{a _ n}^2 -16{b _ n}^2 = 16c _ n\n\\end{align}\\]\r\n\\(c _ 1 = 5^2 -3^2 = 16\\) \u306a\u306e\u3067\r\n\\[\r\nc _ n =\\underline{16^n}\n\\]\r\n (2)<\/strong><\/p>\r\n \\[\r\na _ {k+\\ell} = a _ k a _ {\\ell} +b _ k b _ {\\ell} , \\ b _ {k+\\ell} =b _ k a _ {\\ell} +a _ k b _ {\\ell} \\quad ... [\\text{P}]\r\n\\]\r\n\u3068\u304a\u304f.<\/p>\r\n 1*<\/strong>\u3000\\(\\ell =1\\) \u306e\u3068\u304d 2*<\/strong>\u3000\\(\\ell =m \\ ( m \\geqq 1 )\\) \u306e\u3068\u304d 1*<\/strong> 2*<\/strong> \u3088\u308a, \u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3088\u308a\u984c\u610f\u306f\u793a\u3055\u308c\u305f.<\/p>\r\n (3)<\/strong><\/p>\r\n \\[\\begin{align}\r\nb _ {2\\ell +k} & -c _ {\\ell} b _ k \\\\\r\n& = b _ {\\ell} a _ {\\ell +k} +a _ {\\ell} b _ {\\ell +k} -\\left( {a _ {\\ell}}^2 -{b _ {\\ell}}^2 \\right) b _ k \\\\\r\n& = b _ {\\ell} \\left( a _ k a _ {\\ell} +b _ k b _ {\\ell} \\right) +a _ {\\ell} \\left( b _ k a _ {\\ell} +a _ k b _ {\\ell} \\right) -\\left( {a _ {\\ell}}^2 -{b _ {\\ell}}^2 \\right) b _ k \\\\\r\n& = 2 \\left( a _ k a _ {\\ell} +b _ k b _ {\\ell} \\right) b _ {\\ell} \\\\\r\n& = 2 a _ {\\ell +k} b _ {\\ell}\n\\end{align}\\]\r\n\u3088\u3063\u3066, \\(n =\\ell +k\\) \u3068\u304a\u3051\u3070\r\n\\[\r\nb _ {n+\\ell} -c _ {\\ell} b _ {n-\\ell} = 2 a _ n b _ {\\ell}\n\\]\r\n (4)<\/strong><\/p>\r\n (3)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\r\nb _ {2m+1} -c _ 1 b _ {2m-1} = 2a _ {2m} b _ 1\n\\]\r\n\u4e21\u8fba\u306b \\(c _ {n-m} =16^{n-m}\\) \u3092\u304b\u3051\u308b\u3068\r\n\\[\\begin{align}\r\nc _ {n-m} b _ {2m+1} -c _ {n-m+1} b _ {2m-1} & = 2 c _ {n-m} a _ {2m} b _ 1 \\\\\r\n\\text{\u2234} \\quad c _ {n-m} b _ {2m+1} -c _ {n-(m-1)} b _ {2(m-1)+1} & = 2 c _ {n-m} a _ {2m} b _ 1\n\\end{align}\\]\r\n\u3053\u308c\u306b, \\(m=1 , 2 , \\cdots , n\\) \u3092\u4ee3\u5165\u3057\u305f\u3082\u306e\u3092\u8fba\u3005\u52a0\u3048\u308b\u3068,\r\n\u5de6\u8fba\u306e\u9805\u306f\u76f8\u6bba\u3055\u308c\u3066\r\n\\[\\begin{align}\r\nb _ {2n+1} -c _ n b _ 1 & = 2 b _ 1 \\textstyle\\sum\\limits _ {m=1}^n c _ {n-m} a _ {2m} \\\\\r\n\\text{\u2234} \\quad a _ {2n} +\\textstyle\\sum\\limits _ {m=1}^{n-1} c _ {n-m} a _ {2m} & = \\dfrac{b _ {2n+1}}{2 b _ 1} -\\dfrac{c _ n}{2}\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u6570\u5217 \\(\\{ a _ n \\} , \\{ b _ n \\}\\) \u3092\u6b21\u306e\u3088\u3046\u306b\u5b9a\u7fa9\u3059\u308b. \\[ \\left\\{ \\begin{array}{ll} a _ 1 =5 , \\ b _ 1 =3 , & \\\\ \\left( … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
\r\n
\r\n\u6761\u4ef6\u3088\u308a\r\n\\[\\begin{align}\r\na _ {k+1} & = 5a _ k +3b _ k = a _ k a _ 1 +b _ k b _ 1 , \\\\\r\nb _ {k+1} & = 3a _ k +5b _ k = a _ k b _ 1 +b _ k a _ 1 \\\\\n\\end{align}\\]\r\n\u306a\u306e\u3067, [P] \u304c\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n
\r\n[P] \u304c\u6210\u7acb\u3059\u308b\u3068\u4eee\u5b9a\u3059\u308b\u3068\r\n\\[\\begin{align}\r\na _ {k+m+1} & = 5a _ {k+m} +3b _ {k+m} \\\\\r\n& = 5 a _ k a _ m +5 b _ k b _ m +3 b _ k a _ m +3 a _ k b _ m \\\\\r\n& = a _ k \\left( 5a _ m +3b _ m \\right) +b _ k \\left( 3a _ m +5b _ m \\right) \\\\\r\n& = a _ k a _ {m+1} +b _ k b _ {m+1} , \\\\\r\nb _ {k+m+1} & = 3a _ {k+m} +5b _ {k+m} \\\\\r\n& = 3 a _ k a _ m +3 b _ k b _ m +5 b _ k a _ m +5 a _ k b _ m \\\\\r\n& = a _ k \\left( 3a _ m +5b _ m \\right) +b _ k \\left( 5a _ m +3b _ m \\right) \\\\\r\n& = a _ k b _ {m+1} +b _ k a _ {m+1}\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(\\ell =m+1\\) \u306e\u3068\u304d\u3082 [P] \u304c\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n<\/ol>\r\n