{"id":2128,"date":"2025-07-31T12:23:46","date_gmt":"2025-07-31T03:23:46","guid":{"rendered":"https:\/\/math-friend.com\/?p=2128"},"modified":"2025-08-01T12:57:46","modified_gmt":"2025-08-01T03:57:46","slug":"%e3%80%90%e5%b2%90%e9%98%9c%e5%a4%a7%e5%ad%a6%e5%85%a5%e8%a9%a6%e3%80%91%e6%bc%b8%e5%8c%96%e5%bc%8f%e3%81%a8%e5%af%be%e6%95%b0%e3%81%8b%e3%82%89%e6%95%b0%e5%88%97%e3%81%ae%e4%b8%80%e8%88%ac%e9%a0%85","status":"publish","type":"post","link":"https:\/\/math-friend.com\/?p=2128","title":{"rendered":"\u3010\u5c90\u961c\u5927\u5b66\u5165\u8a66\u3011\u6f38\u5316\u5f0f\u3068\u5bfe\u6570\u304b\u3089\u6570\u5217\u306e\u4e00\u822c\u9805\u3092\u6c42\u3081\u308b\u554f\u984c(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

\u6570\u5217\\(\\{a_n\\}\\), \\(\\{b_n\\}\\)\u3092
$$
\\begin{align}
a_1&=4,\\,\\,a_{n+1}=2^{n+1}\\cdot \\sqrt{a_n}\\,\\,\\,(n=1,2,3,\\cdots)\\\\[1.5ex]
b_n&=\\log_2{a_n}\\,\\,\\,(n=1,2,3,\\cdots)
\\end{align}
$$\u3067\u5b9a\u7fa9\u3059\u308b\u3068\u304d, \u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.

(1) \\(a_2\\), \\(b_1\\), \\(b_2\\)\u3092\u6c42\u3081\u3088.
(2) \\(b_{n+1}\\)\u3092\\(b_n\\)\u3068\\(n\\)\u306e\u5f0f\u3067\u8868\u305b.
(3) \\(c_n=b_{n+1}-b_n\\)\u3068\u3057\u3066, \u6570\u5217\\(\\{c_n\\}\\)\u3092\u5b9a\u7fa9\u3059\u308b\u3068\u304d, \\(c_n\\)\u3092\\(n\\)\u306e\u5f0f\u3067\u8868\u305b.
(4) \\(b_n\\)\u3092\\(n\\)\u306e\u5f0f\u3067\u8868\u305b.
(5) \\(P_n=a_1a_2\\cdots a_n\\)\u3068\u3059\u308b\u3068\u304d, \\(\\log_2{P_n}\\)\u3092\\(n\\)\u306e\u5f0f\u3067\u8868\u305b.
(2025 \u5c90\u961c\u5927\u5b66 [2])<\/span><\/p>\n\n\n\n

\u5bfe\u6570\u306e\u6027\u8cea, \u968e\u5dee\u6570\u5217, \u7b49\u6bd4\u6570\u5217\u306e\u548c\u306e\u516c\u5f0f\u3092\u3075\u3093\u3060\u3093\u306b\u4f7f\u3063\u3066\u89e3\u3044\u3066\u3044\u304f\u554f\u984c\u3067\u3059\u304c, \u4e01\u5be7\u306a\u8a98\u5c0e\u304c\u3042\u308b\u306e\u3067, \u96e3\u3057\u304f\u306f\u3042\u308a\u307e\u305b\u3093.
\u305d\u308c\u3067\u306f\u89e3\u3044\u3066\u3044\u304d\u307e\u3057\u3087\u3046.<\/p>\n\n\n\n

\n
\n

(1) \u6570\u5217\\(\\{a_n\\}\\)\u306e\u6f38\u5316\u5f0f\u3067\\(n=1\\)\u3068\u3057\u3066,
$$
a_2=2^2\\sqrt{a_1}=4\\cdot\\sqrt{4}=8
$$\u3068\u306a\u308a, \\(b_n\\)\u306e\u5b9a\u7fa9\u304b\u3089,
$$
\\begin{align}
b_1&=\\log_2{a_1}=\\log_2{4}=2\\\\[1.5ex]
b_2&=\\log_2{a_2}=\\log_2{8}=3
\\end{align}
$$\u3068\u308f\u304b\u308b.<\/p>\n\n\n\n

(2) \u4ee5\u4e0b\u306e\u3088\u3046\u306b\u5909\u5f62\u3059\u308b\u3053\u3068\u3067\u6c42\u307e\u308b.
$$
\\begin{align}
b_{n+1}&=\\log_2{a_{n+1}}\\\\[1.5ex]
&=\\log_2{\\left(2^{n+1}\\cdot\\sqrt{a_n}\\right)}\\\\[1.5ex]
&=\\log_2{2^{n+1}}+\\log_2{\\sqrt{a_n}}\\\\[1.5ex]
&=n+1+\\frac{1}{2}\\log_2{a_n}\\\\[1.5ex]
&=n+1+\\frac{1}{2}b_n
\\end{align}
$$<\/p>\n\n\n\n

(3) (2)\u3067\u5f97\u3089\u308c\u305f\u5f0f\u306b\u3066\\(n\\)\u306b\\(n+1\\)\u3092\u4ee3\u5165\u3059\u308b\u3068,
$$
b_{n+2}=n+2+\\frac{1}{2}b_{n+1}
$$\u3068\u306a\u308a, \u3053\u306e\u4e21\u8fba\u304b\u3089\\(\\displaystyle b_{n+1}=n+1+\\frac{1}{2}b_n\\)\u3092\u5f15\u304f\u3068,
$$
\\begin{align}
b_{n+2}-b_{n+1}&=1+\\frac{1}{2}b_{n+1}-\\frac{1}{2}b_n\\\\[1.5ex]
&=1+\\frac{1}{2}\\left( b_{n+1}-b_n \\right)
\\end{align}
$$\u3068\u306a\u308a, \\(c_{n+1}=b_{n+2}-b_{n+1}\\)\u306a\u306e\u3067,
$$
c_{n+1}=\\frac{1}{2}c_n+1
$$\u3068\u306a\u308b. \u3053\u308c\u3092\u5909\u5f62\u3059\u308b\u3068,
$$
c_{n+1}-2=\\frac{1}{2}\\left(c_n-2\\right)
$$\u3068\u306a\u308a, \u6570\u5217\\(\\{c_n-2\\}\\)\u306f\u521d\u9805\\(c_1-2\\), \u516c\u6bd4\\(\\displaystyle \\frac{1}{2}\\)\u306e\u7b49\u6bd4\u6570\u5217\u306b\u306a\u308b\u3053\u3068\u304c\u308f\u304b\u308b.
$$
c_1-2=b_2-b_1-2=3-2-2=-1
$$\u3088\u308a,
$$
c_n-2=-\\left(\\frac{1}{2}\\right)^{n-1}
$$\u3067\u3042\u308b\u304b\u3089, \u6700\u7d42\u7684\u306b,
$$
c_n=2-\\left(\\frac{1}{2}\\right)^{n-1}
$$\u3068\u306a\u308b.<\/p>\n\n\n\n

(4) \u6570\u5217\\(\\{c_n\\}\\)\u306f\u6570\u5217\\(\\{b_n\\}\\)\u306e\u968e\u5dee\u6570\u5217\u306b\u306a\u3063\u3066\u3044\u308b\u304b\u3089, \\(n\\geq 2\\)\u306e\u3068\u304d,
$$
b_n=b_1+\\sum_{k=1}^{n-1}c_k
$$\u3067\u3042\u308b. \u3053\u308c\u304b\u3089, \\(n\\geq 2\\)\u306e\u3068\u304d,
$$
\\begin{align}
b_n&=b_1+\\sum_{k=1}^{n-1}\\left\\{2-\\left(\\frac{1}{2}\\right)^{k-1}\\right\\}\\\\[1.5ex]
&=2+\\sum_{k=1}^{n-1}2-\\sum_{k=1}^{n-1}\\left(\\frac{1}{2}\\right)^{k-1}\\\\[1.5ex]
&=2+2(n-1)-\\frac{1-\\left(\\frac{1}{2}\\right)^{n-1}}{1-\\frac{1}{2}}\\\\[1.5ex]
&=2n-2+\\left(\\frac{1}{2}\\right)^{n-2}
\\end{align}
$$\u3068\u306a\u308b. \u3053\u3053\u3067, \\(b_1=2\\)\u3088\u308a, \u3053\u306e\u5f0f\u306f\\(n=1\\)\u3067\u3082\u6210\u308a\u7acb\u3064\u3053\u3068\u304c\u308f\u304b\u308b. \u3088\u3063\u3066,
$$
b_n=2n-2+\\left(\\frac{1}{2}\\right)^{n-2}
$$\u3067\u3042\u308b.<\/p>\n\n\n\n

(5) \u5b9a\u7fa9\u304b\u3089\u660e\u3089\u304b\u306b\\(a_n>0\\)\u306a\u306e\u3067, \u4ee5\u4e0b\u306e\u3088\u3046\u306b\u5909\u5f62\u3059\u308b\u3053\u3068\u3067\u6c42\u307e\u308b.
$$
\\begin{align}
\\log_2{P_n}&=\\log_2{\\left(a_1a_2\\cdots a_n\\right)}\\\\[1.5ex]
&=\\sum_{k=1}^n\\log_2{a_k}\\\\[1.5ex]
&=\\sum_{k=1}^nb_k\\\\[1.5ex]
&=\\sum_{k=1}^n{\\left\\{ 2k-2+\\left(\\frac{1}{2}\\right)^{k-2} \\right\\}}\\\\[1.5ex]
&=2\\sum_{k=1}^nk-\\sum_{k=1}^n2+\\sum_{k=1}^n\\left(\\frac{1}{2}\\right)^{k-2}\\\\[1.5ex]
&=2\\cdot\\frac{1}{2}n(n+1)-2n+\\frac{2\\left\\{1-\\left(\\frac{1}{2}\\right)^n\\right\\}}{1-\\frac{1}{2}}\\\\[1.5ex]
&=n(n+1)-2n+4-\\left(\\frac{1}{2}\\right)^{n-2}\\\\[1.5ex]
&=n^2-n+4-\\left(\\frac{1}{2}\\right)^{n-2}
\\end{align}
$$<\/p>\n<\/div><\/div>\n<\/div><\/div>\n\n\n\n

\u9014\u4e2d, \u7b49\u6bd4\u6570\u5217\u306e\u548c\u306e\u516c\u5f0f\u3092\u4f7f\u3063\u3066\u3044\u308b\u306e\u3067\u88dc\u8db3\u3057\u307e\u3059.

\u521d\u9805\\(a\\), \u516c\u6bd4\\(r\\neq 1\\)\u306e\u7b49\u6bd4\u6570\u5217\u306b\u304a\u3044\u3066, \u521d\u9805\u304b\u3089\u7b2c\\(n\\)\u9805\u307e\u3067\u306e\u548c\u306f,
$$
\\sum_{k=1}^nar^{k-1}=\\frac{a\\left(1-r^n\\right)}{1-r}
$$ \u3068\u306a\u308a\u307e\u3059\u304c, (4), (5)\u3067\u306f\u3053\u308c\u3092\u7528\u3044\u3066\u3044\u307e\u3059.

\u307e\u305a, (4)\u3067\u306f,
$$
\\sum_{k=1}^{n-1}\\left(\\frac{1}{2}\\right)^{k-1}
$$\u306e\u5f62\u306e\u548c\u304c\u51fa\u3066\u304d\u3066\u3044\u307e\u3059\u304c, \u3053\u308c\u306f\u521d\u9805\\(1\\), \u516c\u6bd4\\(\\displaystyle\\frac{1}{2}\\)\u306e\u7b49\u6bd4\u6570\u5217\u306e\u521d\u9805\u304b\u3089\u7b2c\\((n-1)\\)\u9805\u307e\u3067\u306e\u548c\u306a\u306e\u3067,
$$
\\sum_{k=1}^{n-1}\\left(\\frac{1}{2}\\right)^{k-1}=\\frac{1-\\left(\\frac{1}{2}\\right)^{n-1}}{1-\\frac{1}{2}}
$$\u3068\u3057\u3066\u3044\u307e\u3059.

\u307e\u305f, (5)\u3067\u306f,
$$
\\sum_{k=1}^{n}\\left(\\frac{1}{2}\\right)^{k-2}
$$\u306e\u5f62\u306e\u548c\u304c\u51fa\u3066\u304d\u3066\u3044\u307e\u3059\u304c, \u3053\u308c\u306f\u521d\u9805\\(2\\), \u516c\u6bd4\\(\\displaystyle\\frac{1}{2}\\)\u306e\u7b49\u6bd4\u6570\u5217\u306e\u521d\u9805\u304b\u3089\u7b2c\\(n\\)\u9805\u307e\u3067\u306e\u548c\u306a\u306e\u3067,
$$
\\sum_{k=1}^{n}\\left(\\frac{1}{2}\\right)^{k-2}=\\frac{2\\left\\{1-\\left(\\frac{1}{2}\\right)^{n}\\right\\}}{1-\\frac{1}{2}}
$$\u3068\u3057\u3066\u3044\u307e\u3059.<\/p>\n\n\n\n

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