{"id":2668,"date":"2025-08-11T14:00:00","date_gmt":"2025-08-11T05:00:00","guid":{"rendered":"https:\/\/math-friend.com\/?p=2668"},"modified":"2025-08-13T01:28:00","modified_gmt":"2025-08-12T16:28:00","slug":"%e3%80%90%e9%9b%bb%e6%b0%97%e9%80%9a%e4%bf%a1%e5%a4%a7%e5%ad%a6%e3%80%912025","status":"publish","type":"post","link":"https:\/\/math-friend.com\/?p=2668","title":{"rendered":"\u3010\u96fb\u6c17\u901a\u4fe1\u5927\u5b66\u5165\u8a66\u3011\u4ea4\u5dee\u3059\u308b3\u6570\u5217\u306e\u4e00\u822c\u9805\u3068\u7121\u9650\u7d1a\u6570\uff082025\uff09"},"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

\u6b21\u306e\u6761\u4ef6\u306b\u3088\u3063\u3066\u5b9a\u3081\u3089\u308c\u308b\u6570\u5217\\(\\{a_n\\}\\), \\(\\{b_n\\}\\), \\(\\{c_n\\}\\)\u3092\u8003\u3048\u308b.$$
\\begin{aligned}
&a_1=1,\\,\\,\\,b_1=2,\\,\\,\\,c_1=3\\\\[1.5ex]
&a_{n+1}=b_n+c_n,\\\\[1.5ex]
&b_{n+1}=c_n+a_n,\\\\[1.5ex]
&c_{n+1}=a_n+b_n,\\,\\,\\,(n=1,2,3,\\cdots)
\\end{aligned}
$$\u3053\u306e\u3068\u304d, \u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.

(1) \\(a_2\\), \\(b_2\\), \\(c_2\\)\u3092\u6c42\u3081\u3088.
(2) \\(a_n-b_n\\), \\(c_n-b_n\\)\u3092\u305d\u308c\u305e\u308c\\(n\\)\u306e\u5f0f\u3067\u8868\u305b..
(3) \u6570\u5217\\({b_n}\\)\u306e\u4e00\u822c\u9805\u3092\u6c42\u3081\u3088.
(4) \u6570\u5217\\(\\{a_n\\}\\)\u3068\u6570\u5217\\(\\{c_n\\}\\)\u306e\u4e00\u822c\u9805\u3092\u6c42\u3081\u3088.
(5) \\(\\displaystyle \\sum_{n=1}^\\infty\\frac{a_n}{c_n}\\)\u306e\u53ce\u675f, \u767a\u6563\u306b\u3064\u3044\u3066\u8abf\u3079, \u53ce\u675f\u3059\u308b\u3068\u304d\u306f\u305d\u306e\u5024\u3092\u6c42\u3081\u3088.
(2025 \u96fb\u6c17\u901a\u4fe1\u5927\u5b66 [4])<\/span><\/p>\n\n\n\n

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

\n

(1) \u4e0e\u3048\u3089\u308c\u305f\u6f38\u5316\u5f0f\u3067\\(n=1\\)\u3068\u3059\u308b\u3053\u3068\u3067,$$
\\begin{align}
a_2=b_1+c_1=2+3=5\\\\[1.5ex]
b_2=c_1+a_1=3+1=4\\\\[1.5ex]
c_2=a_1+b_1=1+2=3\\\\[1.5ex]
\\end{align}$$\u3068\u6c42\u307e\u308b.<\/p>\n\n\n\n

(2) \u6f38\u5316\u5f0f\u304b\u3089\\(a_{n+1}-b_{n+1}\\)\u3092\u8a08\u7b97\u3059\u308b\u3068,$$
\\begin{align}
a_{n+1}-b_{n+1}&=(b_n+c_n)-(c_n-a_n)\\\\[1.5ex]
&=-(a_n-b_n)
\\end{align}
$$\u3068\u306a\u308a, \u6570\u5217\\(\\{a_n-b_n\\}\\)\u306f\u516c\u6bd4\u304c\\(-1\\)\u306e\u7b49\u6bd4\u6570\u5217\u3068\u306a\u308b. \u3088\u3063\u3066,
$$
a_n-b_n=(a_1-b_1)\\cdot(-1)^{n-1}=(-1)^n
$$\u3068\u6c42\u307e\u308b.

\u540c\u69d8\u306b, \\(c_{n+1}-b_{n+1}\\)\u3092\u8a08\u7b97\u3059\u308b\u3068,$$
\\begin{align}
c_{n+1}-b_{n+1}&=(a_n+b_n)-(c_n-a_n)\\\\[1.5ex]
&=-(c_n-b_n)
\\end{align}
$$\u3068\u306a\u308a, \u6570\u5217\\(\\{c_n-b_n\\}\\)\u3082\u516c\u6bd4\u304c\\(-1\\)\u306e\u7b49\u6bd4\u6570\u5217\u3068\u306a\u308b. \u3088\u3063\u3066,
$$
c_n-b_n=(c_1-b_1)\\cdot(-1)^{n-1}=(-1)^{n-1}
$$\u3068\u6c42\u307e\u308b.<\/p>\n\n\n\n

(3) (2)\u3067\u6c42\u3081\u305f\u4ee5\u4e0b\u306e2\u3064\u306e\u95a2\u4fc2\u5f0f\u306b\u3064\u3044\u3066,$$
\\begin{align}
a_n-b_n&=(-1)^n\\\\[1.5ex]
c_n-b_n&=(-1)^{n-1}
\\end{align}
$$\u3053\u306e\u8fba\u3005\u3092\u8db3\u3059\u3068, \\(a_n+c_n=b_{n+1}\\)\u3088\u308a,$$
b_{n+1}-2b_n=(-1)^n+(-1)^{n-1}
$$\u3068\u306a\u308b. \u3053\u3053\u3067, \\((-1)^n=-(-1)^{n-1}\\)\u3088\u308a, \u3053\u306e\u5de6\u8fba\u306f\\(0\\)\u306b\u306a\u308b\u304b\u3089,$$
b_{n+1}=2b_n
$$\u304c\u308f\u304b\u308b. \u3088\u3063\u3066, \u6570\u5217\\(\\{b_n\\}\\)\u306f\u516c\u6bd4\u304c\\(2\\)\u306e\u7b49\u6bd4\u6570\u5217\u3068\u306a\u308a, \\(b_1=2\\)\u3088\u308a,$$
b_n=b_1\\cdot 2^{n-1}=2^n
$$\u3068\u3057\u3066, \\(b_n\\)\u306e\u4e00\u822c\u9805\u304c\u6c42\u307e\u3063\u305f.<\/p>\n\n\n\n

(4) (2), (3)\u306e\u7d50\u679c\u304b\u3089,$$
\\begin{align}
a_n=b_n+(-1)^n=2^n+(-1)^n\\\\[1.5ex]
c_n=b_n+(-1)^{n-1}=2^n+(-1)^{n-1}
\\end{align}
$$\u3068\u6c42\u307e\u308b.<\/p>\n\n\n\n

(5) \\(\\displaystyle \\frac{a_n}{c_n}\\)\u306e\u6975\u9650\u3092\u6c42\u3081\u308b\u3068,$$
\\begin{align}
\\lim_{n\\rightarrow \\infty}\\frac{a_n}{c_n}&=\\lim_{n\\rightarrow \\infty}\\frac{2^n+(-1)^n}{2^n+(-1)^{n-1}}\\\\[1.5ex]
&=\\lim_{n\\rightarrow \\infty}\\frac{1+(-\\frac{1}{2})^n}{1+\\frac{1}{2}\\cdot(-\\frac{1}{2})^{n-1}}\\\\[1.5ex]
&=1
\\end{align}
$$\u3068\u306a\u308a, \u3053\u308c\u306f\\(0\\)\u306b\u53ce\u675f\u3057\u306a\u3044. \u3088\u3063\u3066, \\(\\displaystyle\\sum_{n=1}^\\infty\\frac{a_n}{c_n}\\)\u3082\u53ce\u675f\u3057\u306a\u3044\u3053\u3068\u304c\u308f\u304b\u308b.<\/p>\n<\/div><\/div>\n\n\n\n

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