{"id":1190,"date":"2025-07-14T13:36:57","date_gmt":"2025-07-14T04:36:57","guid":{"rendered":"https:\/\/math-friend.com\/?p=1190"},"modified":"2025-08-01T09:48:10","modified_gmt":"2025-08-01T00:48:10","slug":"%e3%80%90%e5%8c%97%e6%b5%b7%e9%81%93%e5%a4%a7%e5%ad%a6%e5%85%a5%e8%a9%a6%e3%80%91%e9%9a%a3%e6%8e%a53%e9%a0%85%e9%96%93%e3%81%ae%e6%95%b0%e5%88%97%e3%81%a8%e3%81%9d%e3%81%ae%e9%80%86%e6%95%b0%e3%81%ae","status":"publish","type":"post","link":"https:\/\/math-friend.com\/?p=1190","title":{"rendered":"\u3010\u5317\u6d77\u9053\u5927\u5b66\u5165\u8a66\u3011\u96a3\u63a53\u9805\u9593\u306e\u6570\u5217\u3068\u305d\u306e\u9006\u6570\u306e\u548c(2025)"},"content":{"rendered":"\n

\u4eca\u56de\u306f\u3053\u3061\u3089\u306e\u554f\u984c\u3092\u89e3\u3044\u3066\u3044\u304d\u307e\u3059.<\/p>\n\n\n\n

\u4ee5\u4e0b\u306e\u6f38\u5316\u5f0f\u3067\u4e0e\u3048\u3089\u308c\u308b\u6570\u5217\\(\\{a_n\\}\\)\u306f\\(a_1=1\\), \\(a_2=3\\)\u3067\u4ee5\u4e0b\u306e\u6f38\u5316\u5f0f\u3092\u6e80\u305f\u3059.
$$
(n+1)a_{n+2}-(2n+3)a_{n+1}+(n+2)a_n=0\\,\\,(n=1,2,3,\\cdots)
$$\u307e\u305f, \\(b_n=a_{n+1}-a_n\\)\u306b\u3088\u308a, \u6570\u5217\\(\\{b_n\\}\\)\u3092\u5b9a\u3081\u308b.
\u3053\u306e\u3068\u304d, \u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.

(1) \\(\\displaystyle b_{n+1}=\\frac{n+2}{n+1}b_n\\)\u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u305b.

(2) \\(\\{a_n\\}\\)\u306e\u4e00\u822c\u9805\u3092\u6c42\u3081\u3088.

(3) \\(\\displaystyle \\sum_{n=1}^{225}=\\frac{1}{a_n}\\)\u3092\u6c42\u3081\u3088.
(2025 \u5317\u6d77\u9053\u5927\u5b66\u6587\u7cfb[3])<\/span><\/p>\n\n\n\n

\u96a3\u63a53\u9805\u9593\u306e\u6f38\u5316\u5f0f\u3068\u3044\u3046\u3053\u3068\u3067, \u96e3\u3057\u305d\u3046\u306b\u611f\u3058\u307e\u3059\u304c, \u4e01\u5be7\u306a\u8a98\u5c0e\u304c\u3042\u308b\u306e\u3067\u662f\u975e\u3068\u3082\u89e3\u304d\u305f\u3044\u3068\u3053\u308d\u3067\u3059.

\u305d\u308c\u3067\u306f\u89e3\u3044\u3066\u3044\u304d\u307e\u3057\u3087\u3046.<\/p>\n\n\n\n

\n

(1) \\(b_n\\)\u306e\u5b9a\u7fa9\u304b\u3089
$$
a_{n+1}=b_n+a_n
$$\u3067\u3042\u308a, \u540c\u69d8\u306b,
$$
a_{n+2}=b_{n+1}+a_{n+1} = b_{n+1}+b_n+a_n
$$\u3068\u306a\u308b. \u3053\u308c\u3092\\(\\{a_n\\}\\)\u306e\u6f38\u5316\u5f0f\u306b\u4ee3\u5165\u3059\u308b\u3068,
$$
(n+1)(b_{n+1}+b_n+a_n)-(2n+3)(b_n+a_n)+(n+2)a_n=0
$$\u3068\u306a\u308a, \u3053\u308c\u3092\u5c55\u958b\u3059\u308b\u3068\\(a_n\\)\u304c\u6d88\u3048, \u4ee5\u4e0b\u306e\\(\\{b_n\\}\\)\u306e\u95a2\u4fc2\u5f0f\u304c\u5f97\u3089\u308c\u308b.
$$
(n+1)b_{n+1}-(n+2)b_n=0
$$\u7b2c2\u9805\u3092\u79fb\u9805\u3057, \u4e21\u8fba\u3092\\(n+1\\)\u3067\u5272\u308b\u3053\u3068\u3067, \u793a\u3059\u3079\u304d\u6f38\u5316\u5f0f\u304c\u5f97\u3089\u308c\u308b.
$$
b_{n+1}=\\frac{n+2}{n+1}b_n
$$<\/p>\n\n\n\n

(2) (1)\u306e\u6f38\u5316\u5f0f\u3092\u7d9a\u3051\u3066\u7528\u3044\u308b\u3053\u3068\u3067\\(b_n\\)\u306f\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u8a08\u7b97\u304c\u3067\u304d\u308b.
$$
\\begin{align}
b_n&=\\frac{n+1}{n}b_{n-1}\\\\[1.5ex]
&=\\frac{n+1}{n}\\frac{n}{n-1}b_{n-2}\\\\[1.5ex]
&=\\frac{n+1}{n}\\frac{n}{n-1}\\frac{n-1}{n-2}b_{n-3}\\\\[1.5ex]
&\\cdots\\\\[1.5ex]
&=\\frac{n+1}{n}\\frac{n}{n-1}\\frac{n-1}{n-2}\\cdots\\frac{3}{2}b_1\\\\[1.5ex]
&=\\frac{n+1}{\\cancel{n}}\\frac{\\cancel{n}}{\\cancel{n-1}}\\frac{\\cancel{n-1}}{\\cancel{n-2}}\\cdots\\frac{\\cancel{3}}{2}b_1\\\\[1.5ex]
&=\\frac{n+1}{2}b_1
\\end{align}
$$
\u3053\u3053\u3067,
$$
b_1=a_2-a_1=3-1=2
$$\u3088\u308a,
$$
b_n=\\frac{n+1}{2}\\cdot 2=n+1
$$
\u3068\u6c42\u307e\u308b. \\(b_n\\)\u306f\\(a_n\\)\u306e\u968e\u5dee\u6570\u5217\u306b\u306a\u3063\u3066\u3044\u308b\u3053\u3068\u304b\u3089, \\(n\\geq 2\\)\u306e\u3068\u304d,
$$
\\begin{align}
a_n&=a_1+\\sum_{k=1}^{n-1}b_k=1+\\sum_{k=1}^{n-1}(k+1)\\\\[1.5ex]
&=1+\\sum_{k=1}^{n-1}(k+1)=1+\\frac{n(n-1)}{2}+(n-1)\\\\[1.5ex]
&=1+\\frac{n^2}{2}-\\frac{n}{2}+n-1=\\frac{n^2}{2}+\\frac{n}{2}=\\frac{n(n+1)}{2}
\\end{align}
$$\u3068\u306a\u308b. \u3053\u308c\u306f\\(n=1\\)\u306e\u3068\u304d\u3082\u6210\u308a\u7acb\u3064\u306e\u3067, \u5168\u3066\u306e\u81ea\u7136\u6570\\(n\\)\u306b\u5bfe\u3057\u3066,
$$
a_n=\\frac{n(n+1)}{2}
$$
\u3067\u3042\u308b\u3053\u3068\u304c\u308f\u304b\u308b.<\/p>\n\n\n\n

(3) \\(\\displaystyle \\frac{1}{a_n}\\)\u306f\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u90e8\u5206\u5206\u6570\u5206\u89e3\u3067\u304d\u308b.
$$
\\frac{1}{a_n}=\\frac{2}{n(n+1)}=2\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right)
$$ \u3053\u308c\u304b\u3089,
$$
\\begin{align}
\\sum_{n=1}^{225}\\frac{1}{a_n}&=2\\sum_{n=1}^{225}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right)\\\\[1.5ex]
&=2\\left\\{\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\cdots +\\left(\\frac{1}{225}-\\frac{1}{226}\\right)\\right\\}\\\\[1.5ex]
&=2\\left\\{\\left(1-\\cancel{\\frac{1}{2}}\\right)+\\left(\\cancel{\\frac{1}{2}}-\\cancel{\\frac{1}{3}}\\right)+\\cdots +\\left(\\cancel{\\frac{1}{225}}-\\frac{1}{226}\\right)\\right\\}\\\\[1.5ex]
&=2\\left(1-\\frac{1}{226}\\right)=\\frac{225}{113}
\\end{align}
$$\u3068\u6c42\u307e\u308b.<\/p>\n<\/div><\/div>\n\n\n\n

(1)\u306f\\(\\{a_n\\}\\)\u306e\u6f38\u5316\u5f0f\u3092\\(b_n\\)\u306e\u6f38\u5316\u5f0f\u306b\u7f6e\u304d\u63db\u3048\u308b\u3068\u3044\u3046\u65b9\u91dd\u3067\u8003\u3048\u308c\u3070, \\(a_{n+2}\\), \\(a_{n+1}\\)\u3092\u6d88\u3057\u3066\u3044\u3053\u3046\u3068\u3044\u3046\u767a\u60f3\u306b\u884c\u304d\u7740\u304f\u3068\u601d\u3044\u307e\u3059. \u4e0e\u3048\u3089\u308c\u3066\u3044\u308b\u5f0f\u304c\u5c11\u306a\u3044\u3053\u3068\u3082\u3042\u308a, \u305d\u308c\u4ee5\u5916\u306b\u89e3\u6cd5\u304c\u601d\u3044\u3064\u304d\u307e\u305b\u3093.

(2)\u306f(1)\u306e\u6f38\u5316\u5f0f\u304b\u3089\\(b_n\\)\u304c\u6c42\u307e\u308a\u305d\u3046\u3067\u3042\u308b\u3053\u3068, \u307e\u305f\\(\\{b_n\\}\\)\u304c\\(\\{a_n\\}\\)\u968e\u5dee\u6570\u5217\u3067\u3042\u308b\u3068\u3044\u3046\u3053\u3068\u306b\u6c17\u3065\u3051\u3070\u96e3\u3057\u304f\u306a\u3044\u3067\u3057\u3087\u3046.

(3)\u306f\\(\\displaystyle \\frac{1}{n(n+1)}\\)\u306e\u5f62\u3092\u3057\u305f\u5206\u6570\u306e\u548c\u3092\u6c42\u3081\u308b\u969b\u306f, \u90e8\u5206\u5206\u6570\u5206\u89e3\u3092\u884c\u3046\u3068\u3044\u3046\u5b9a\u77f3\u3092\u77e5\u3063\u3066\u3044\u308c\u3070\u554f\u984c\u306a\u304f\u89e3\u3051\u308b\u306f\u305a\u3067\u3059.

\u90e8\u5206\u5206\u6570\u5206\u89e3\u306b\u95a2\u3057\u3066\u306f, \u4e00\u822c\u306b\u4ee5\u4e0b\u306e\u516c\u5f0f\u3092\u899a\u3048\u3066\u304a\u304f\u3068\u4fbf\u5229\u3067\u3059.
$$
\\begin{align}
\\frac{1}{x(x+a)}&=\\frac{1}{a}\\left(\\frac{1}{x}-\\frac{1}{x+a}\\right) \\,\\, (a\\neq 0),\\\\
\\frac{1}{(x+a)(x+b)}&=\\frac{1}{b-a}\\left(\\frac{1}{x+b}-\\frac{1}{x+a}\\right) \\,\\, (a\\neq b)
\\end{align}
$$\u4eca\u56de\u306e\u554f\u984c\u306e\u90e8\u5206\u5206\u6570\u5206\u89e3\u306f2\u756a\u76ee\u306e\u5f0f\u3067\\(x=n\\), \\(a=0\\), \\(b=1\\)\u3068\u3057\u305f\u3082\u306e\u3067\u3059.<\/p>\n\n\n\n

youtube\u3067\u3082\u89e3\u8aac\u3057\u3066\u3044\u307e\u3059.<\/p>\n\n\n\n