\u6570\u5217 \\(\\{ a _ n \\}\\) \u3092\r\n\\[\r\na _ 1 = 5 , \\ a _ {n+1} = \\dfrac{4 a _ n -9}{a _ n -2} \\quad ( n = 1, 2, 3, \\cdots )\r\n\\]\r\n\u3067\u5b9a\u3081\u308b. \u307e\u305f\u6570\u5217 \\(\\{ b _ n \\}\\) \u3092\r\n\\[\r\nb _ n = \\dfrac{a _ 1 +2 a _ 2 +\\cdots +n a _ n}{1 +2 +\\cdots +n} \\quad ( n = 1, 2, 3, \\cdots )\r\n\\]\r\n\u3068\u5b9a\u3081\u308b.<\/p>\r\n
(1)<\/strong>\u3000\u6570\u5217 \\(\\{ a _ n \\}\\) \u306e\u4e00\u822c\u9805\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000\u3059\u3079\u3066\u306e \\(n\\) \u306b\u5bfe\u3057\u3066, \u4e0d\u7b49\u5f0f \\(b _ n \\leqq 3 +\\dfrac{4}{n+1}\\) \u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n (3)<\/strong>\u3000\u6975\u9650\u5024 \\(\\displaystyle\\lim _ {n \\rightarrow \\infty} b _ n\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong>\r\n\\[\r\na _ n = \\dfrac{6n-1}{2n-1} \\quad ... [ \\text{A} ]\r\n\\]\r\n\u3067\u3042\u308b\u3053\u3068\u3092, \u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3092\u7528\u3044\u3066\u793a\u3059.<\/p>\r\n 1*<\/strong>\u3000\\(n = 1\\) \u306e\u3068\u304d 2*<\/strong>\u3000\\(n = k \\ ( k \\geqq 1 )\\) \u306e\u3068\u304d, [A] \u304c\u6210\u7acb\u3059\u308b, \u3059\u306a\u308f\u3061 \u3088\u3063\u3066, \u3059\u3079\u3066\u306e\u81ea\u7136\u6570 \\(n\\) \u306b\u3064\u3044\u3066 [A] \u304c\u6210\u7acb\u3057\r\n\\[\r\na _ n = \\underline{\\dfrac{6n-1}{2n-1}}\r\n\\]\r\n (2)<\/strong><\/p>\r\n \\(S _ n = a _ n +2 a _ n +\\cdots +n a _ n\\) \u3068\u304a\u3044\u3066, \u793a\u3057\u305f\u3044\u4e0d\u7b49\u5f0f\u3092\u5909\u5f62\u3059\u308b\u3068\r\n\\[\\begin{align}\r\n\\dfrac{2 S _ n}{n (n+1)} & \\leqq 3 +\\dfrac{4}{n+1} \\\\\r\n2 S _ n & \\leqq 3n (n+1) +4n \\\\\r\n\\text{\u2234} \\quad 2 S _ n & \\leqq 3n^2 +7n \\quad ... [ \\text{B} ]\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, [B] \u304c\u6210\u7acb\u3059\u308b\u3053\u3068\u3092\u793a\u305b\u3070\u3088\u3044. 1*<\/strong>\u3000\\(n = 1\\) \u306e\u3068\u304d 2*<\/strong>\u3000\\(n = k \\ ( k \\geqq 1 )\\) \u306e\u3068\u304d, [B] \u304c\u6210\u7acb\u3059\u308b, \u3059\u306a\u308f\u3061 \u3088\u3063\u3066, \u3059\u3079\u3066\u306e\u81ea\u7136\u6570 \\(n\\) \u306b\u3064\u3044\u3066 [B] \u304c\u6210\u7acb\u3057, \u984c\u610f\u306f\u793a\u3055\u308c\u305f.<\/p>\r\n (3)<\/strong><\/p>\r\n (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\r\na _ n = 3 +\\dfrac{2}{2n-1} \\gt 3\r\n\\]\r\n\u306a\u306e\u3067\r\n\\[\r\nb _ n \\gt \\dfrac{3 ( 1 +2 +\\cdots +n )}{1 +2 +\\cdots +n} = 3\r\n\\]\r\n (2)<\/strong> \u306e\u7d50\u679c\u3068\u3042\u308f\u305b\u308b\u3068\r\n\\[\r\n3 \\lt b _ n \\leqq \\underline{3 +\\dfrac{4}{n+1}} _ {[1]}\r\n\\]\r\n\u3053\u3053\u3067 \\(\\displaystyle\\lim _ {n \\rightarrow \\infty} [1] = 3\\) \u306a\u306e\u3067\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} b _ n = \\underline{3}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\u6570\u5217 \\(\\{ a _ n \\}\\) \u3092 \\[ a _ 1 = 5 , \\ a _ {n+1} = \\dfrac{4 a _ n -9}{a _ n -2} \\quad ( n = 1, 2, 3, \\cdots ) \\ … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
\r\n
\r\n\\[\r\na _ 1 = \\dfrac{5}{1} = 5\r\n\\]\r\n\u306a\u306e\u3067, [A] \u306f\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n
\r\n\\[\r\na _ k = \\dfrac{6k-1}{2k-1}\r\n\\]\r\n\u3068\u4eee\u5b9a\u3059\u308b\u3068\r\n\\[\\begin{align}\r\na _ {k+1} &= \\dfrac{4 a _ k -9}{a _ k -2} \\\\\r\n& = \\dfrac{4(6k-1) -9(2k-1)}{(6k-1) -2(2k-1)} \\\\\r\n& = \\dfrac{6(k+1) -1}{2(k+1) -1}\r\n\\end{align}\\]\r\n\u3086\u3048\u306b, \\(n = k+1\\) \u306e\u3068\u304d, [A] \u304c\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n<\/ol>\r\n
\r\n\u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3092\u7528\u3044\u3066\u793a\u3059.<\/p>\r\n\r\n
\r\n\\[\r\n2 S _ 1 = 2 \\cdot 5 \\leqq 3+7\r\n\\]\r\n\u306a\u306e\u3067, [B] \u306f\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n
\r\n\\[\r\n2 S _ k \\leqq 3k^2 +7k\r\n\\]\r\n\u3068\u4eee\u5b9a\u3059\u308b\u3068\r\n\\[\\begin{align}\r\n2 S _ {k+1} & = 2 S _ k +2(k+1) a _ {k+1} \\\\\r\n& \\leqq 3k^2 +7k +2(k+1) \\left( 3 +\\dfrac{2}{2k+1} \\right) \\\\\r\n& = 3k^2 +13k +6 +4 \\cdot \\dfrac{k+1}{2k+1} \\\\\r\n& \\lt 3k^2 +13k +10 \\\\\r\n& = 3(k+1)^2 +7(k+1)\r\n\\end{align}\\]\r\n\u3086\u3048\u306b, \\(n = k+1\\) \u306e\u3068\u304d, [B] \u304c\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n<\/ol>\r\n