\u6570\u5217 \\(\\{ a _ n \\}\\) \u306f\r\n\\[\r\na _ 1 = 5 , \\quad {a _ 1}^2 +{a _ 2}^2 + \\cdots +{a _ n}^2 = \\dfrac{2}{3} a _ n a _ {n+1} \\quad ( n = 1, 2, 3, \\cdots )\r\n\\]\r\n\u3092\u307f\u305f\u3059\u3068\u3059\u308b. \u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n
(1)<\/strong>\u3000\\(a _ 2 , a _ 3\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(a _ {n+2}\\) \u3092 \\(a _ n , a _ {n+1}\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/p><\/li>\r\n (3)<\/strong>\u3000\u4e00\u822c\u9805 \\(a _ n\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \u4e0e\u3048\u3089\u308c\u305f\u6f38\u5316\u5f0f\u306b\u3064\u3044\u3066 \\(n = 1\\) \u3068\u3059\u308c\u3070\r\n\\[\\begin{align}\r\n5^2 & = \\dfrac{2}{3} \\cdot 5 a_2 \\\\\r\n\\text{\u2234} \\quad a_2 & = \\underline{\\dfrac{15}{2}}\r\n\\end{align}\\]\r\n\u307e\u305f, \\(n = 2\\) \u3068\u3059\u308c\u3070\r\n\\[\\begin{align}\r\n5^2 +\\left( \\dfrac{15}{2} \\right)^2 & = \\dfrac{2}{3} \\cdot \\dfrac{15}{2} a_3 \\\\\r\n\\text{\u2234} \\quad a_3 & = \\underline{\\dfrac{65}{4}}\r\n\\end{align}\\]\r\n (2)<\/strong><\/p>\r\n \\[\\begin{align}\r\n{a_1}^2 +{a_2}^2 +\\cdots +{a_n}^2 & = \\dfrac{2}{3} a_n a _ {n+1} \\quad ... [1] \\\\\r\n{a_1}^2 +{a_2}^2 +\\cdots +{a_n}^2 +{a _ {n+1}}^2 & = \\dfrac{2}{3} a _ {n+1} a _ {n+2} \\quad ... [2]\r\n\\end{align}\\]\r\n\\([2] -[1]\\) \u3088\u308a\r\n\\[\r\n{a _ {n+1}}^2 = \\dfrac{2}{3} a _ {n+1} \\left( a _ {n+2} -a_n \\right)\r\n\\]\r\n\\(( [1] \\text{\u306e\u5de6\u8fba} ) \\neq 0\\) \u306a\u306e\u3067 \\(a _ {n+1} \\neq 0\\) \u3060\u304b\u3089\r\n\\[\\begin{align}\r\na _ {n+1} & = \\dfrac{2}{3} \\left( a _ {n+2} -a_n \\right) \\\\\r\n\\text{\u2234} \\quad a _ {n+2} & = \\underline{\\dfrac{3}{2} a _ {n+1} +a_n} \\quad ... [3]\r\n\\end{align}\\]\r\n (3)<\/strong><\/p>\r\n[3] \u3092\u5909\u5f62\u3059\u308b\u3068\r\n\\[\r\na _ {n+2} +\\dfrac{1}{2} a _ {n+1} = 2 \\left( a _ {n+1} +\\dfrac{1}{2} a_n \\right)\r\n\\]\r\n\u6570\u5217 \\(\\left\\{ a _ {n+1} +\\dfrac{1}{2} a_n \\right\\}\\) \u306f, \u521d\u9805 \\(a_2 +\\dfrac{1}{2} a_1 = \\dfrac{15}{2} +\\dfrac{5}{2} = 10\\) , \u516c\u6bd4 \\(2\\) \u306e\u7b49\u6bd4\u6570\u5217\u306a\u306e\u3067\r\n\\[\\begin{align}\r\na _ {n+1} +\\dfrac{1}{2} a_n & = 10 \\cdot 2^{n-1} = 5 \\cdot 2^n \\\\\r\n\\text{\u2234} \\quad a _ {n+1} -2^{n+2} & = -\\dfrac{1}{2} \\left( a_n -2^{n+1} \\right)\r\n\\end{align}\\]\r\n\u6570\u5217 \\(\\left\\{ a_n -2^{n+1} \\right\\}\\) \u306f, \u521d\u9805 \\(a_1 -2^2 = 1\\) , \u516c\u6bd4 \\(-\\dfrac{1}{2}\\) \u306e\u7b49\u6bd4\u6570\u5217\u306a\u306e\u3067\r\n\\[\\begin{align}\r\na_n -2^{n+1} & = \\left( -\\dfrac{1}{2} \\right)^{n-1} \\\\\r\n\\text{\u2234} \\quad a_n & = \\underline{2^{n+1} +\\left( -\\dfrac{1}{2} \\right)^{n-1}}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u6570\u5217 \\(\\{ a _ n \\}\\) \u306f \\[ a _ 1 = 5 , \\quad {a _ 1}^2 +{a _ 2}^2 + \\cdots +{a _ n}^2 = \\dfrac{2}{3} a _ n a _ {n … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n