\u6570\u5217 \\(\\{ a _ n \\}\\) \u3092,\r\n\\[\\begin{align}\r\na _ 1 & = 1 , \\\\\r\n(n+3) a _ {n+1} -n a _ n & = \\dfrac{1}{n+1} -\\dfrac{1}{n+2} \\quad ( n = 1 , 2 , 3 , \\cdots )\r\n\\end{align}\\]\r\n\u306b\u3088\u3063\u3066\u5b9a\u3081\u308b.<\/p>\r\n
(1)<\/strong>\u3000\\(b _ n = n(n+1)(n+2) a _ n \\ ( n = 1 , 2 , 3 , \\cdots )\\) \u306b\u3088\u3063\u3066\u5b9a\u307e\u308b\u6570\u5217 \\(\\{ b _ n \\}\\) \u306e\u4e00\u822c\u9805\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000\u7b49\u5f0f\r\n\\[\r\np(n+1)(n+2) +qn(n+2) +rn(n+1) = b _ n \\quad ( n = 1 , 2 , 3 , \\cdots )\r\n\\]\r\n\u6210\u308a\u7acb\u3064\u3088\u3046\u306b, \u5b9a\u6570 \\(p , q , r\\) \u306e\u5024\u3092\u5b9a\u3081\u3088.<\/p><\/li>\r\n (3)<\/strong>\u3000\\(\\textstyle\\sum\\limits _ {k=1}^n a _ k\\) \u3092 \\(n\\) \u306e\u5f0f\u3067\u8868\u305b.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \u6f38\u5316\u5f0f\u306e\u4e21\u8fba\u306b \\((n+1)(n+2)\\) \u3092\u304b\u3051\u308b\u3068\r\n\\[\\begin{align}\r\n(n+1)(n+2)(n+3) a _ {n+1} -n(n+1)(n+2) a _ n & = (n+2) -(n+1) \\\\\r\n\\text{\u2234} \\quad b _ {n+1} -b _ n & = 1\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(\\{ b _ n \\}\\) \u306f, \u521d\u9805 \\(b _ 1 = 1 \\cdot 2 \\cdot 3 a _ 1 = 6\\) , \u516c\u5dee \\(1\\) \u306e\u7b49\u5dee\u6570\u5217\u3067\r\n\\[\r\nb _ n = 6 +1 \\cdot (n-1) = \\underline{n-5}\r\n\\]\r\n (2)<\/strong><\/p>\r\n \\[\\begin{align}\r\np (n+1) & (n+2) +qn(n+2) +rn(n+1) \\\\\r\n& = p (n^2+3n+2) +q (n^2+2n) r (n^2+n) \\\\\r\n& = (p+q+r) n^2 +(3p+2q+r)n +2p\r\n\\end{align}\\]\r\n\u3053\u308c\u3068 (1)<\/strong> \u306e\u7d50\u679c\u3092\u6bd4\u8f03\u3059\u308c\u3070,\r\n\\[\\begin{gather}\r\n\\left\\{ \\begin{array}{l} p+q+r=0 \\\\ 3p+2q+r=1 \\\\ 2p=5 \\end{array} \\right. \\\\\r\n\\text{\u2234} \\quad p =\\underline{\\dfrac{5}{2}} , \\ q = \\underline{-4} , \\ r = \\underline{\\dfrac{3}{2}}\r\n\\end{gather}\\]\r\n (3)<\/strong><\/p>\r\n (2)<\/strong> \u306e\u7d50\u679c\u306b\u3064\u3044\u3066, \u4e21\u8fba\u3092 \\(n(n+1)(n+2)\\) \u3067\u5272\u308c\u3070\r\n\\[\\begin{align}\r\na _ n & = \\dfrac{5}{2n} -\\dfrac{4}{n+1} +\\dfrac{3}{2(n+2)} \\\\\r\n& = \\dfrac{5}{2} \\left( \\dfrac{1}{n} -\\dfrac{1}{n+1} \\right) -\\dfrac{3}{2} \\left( \\dfrac{1}{n+1} -\\dfrac{1}{n+2} \\right)\r\n\\end{align}\\]\r\n\u3086\u3048\u306b\r\n\\[\\begin{align}\r\n\\textstyle\\sum\\limits _ {k=1}^n a _ k & = \\dfrac{5}{2} \\left( 1 -\\dfrac{1}{n+1} \\right) -\\dfrac{3}{2} \\left( \\dfrac{1}{2} -\\dfrac{1}{n+2} \\right) \\\\\r\n& = \\dfrac{7}{4} -\\dfrac{5}{2(n+1)} +\\dfrac{3}{2(n+2)} \\\\\r\n& = \\dfrac{7(n+1)(n+2) -10(n+2) +6(n+1)}{4(n+1)(n+2)} \\\\\r\n& = \\underline{\\dfrac{n(7n+17)}{4(n+1)(n+2)}}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u6570\u5217 \\(\\{ a _ n \\}\\) \u3092, \\[\\begin{align} a _ 1 & = 1 , \\\\ (n+3) a _ {n+1} -n a _ n & = \\dfrac{1}{n+1} -\\dfrac{1}{ … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n