{"id":2823,"date":"2025-08-21T02:10:57","date_gmt":"2025-08-20T17:10:57","guid":{"rendered":"https:\/\/math-friend.com\/?p=2823"},"modified":"2025-08-21T02:11:18","modified_gmt":"2025-08-20T17:11:18","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%e6%95%b0%e5%88%97%e3%81%ae%e5%92%8c%e3%81%8c%e6%9d%a1%e4%bb%b6%e5%bc%8f%e3%81%ab%e5%85%a5%e3%81%a3%e3%81%a6%e3%81%84","status":"publish","type":"post","link":"https:\/\/math-friend.com\/?p=2823","title":{"rendered":"\u3010\u5317\u6d77\u9053\u5927\u5b66\u5165\u8a66\u3011\u90e8\u5206\u548c\u306e\u4e0e\u3048\u3089\u308c\u305f\u6570\u5217\u306e\u4e00\u822c\u9805\u3068\u9006\u6570\u548c\u306e\u8a08\u7b97\uff082021)"},"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

\u521d\u9805\u304b\u3089\u7b2c\\(n\\)\u9805\u307e\u3067\u306e\u548c\\(S_n\\)\u304c,$$
S_n=\\frac{1}{6}n(n+1)(2n+7)\\,\\,\\,(n=1,2,3,\\cdots)
$$\u3067\u4e0e\u3048\u3089\u308c\u308b\u6570\u5217\\(\\{a_n\\}\\)\u304c\u3042\u308b.

(1) \u6570\u5217\\(\\{a_n\\}\\)\u306e\u4e00\u822c\u9805\u3092\u6c42\u3081\u3088.
(2) \\(\\displaystyle \\sum_{k=1}^n\\frac{1}{a_k}\\)\u3092\u6c42\u3081\u3088.
(2021 \u5317\u6d77\u9053\u5927\u5b66 \u6587\u7cfb [1])<\/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) \u307e\u305a\\(a_1\\)\u306f,$$
a_1=S_1=\\frac{1}{6}\\times 1\\times(1+1)\\times(2\\times 1+7)=3
$$\u3068\u6c42\u307e\u308b. \u6b21\u306b\\(n\\geq 2\\)\u306e\u3068\u304d, \\(a_n\\)\u306f,$$
\\begin{align}
a_n&=S_n-S_{n-1}\\\\[1.5ex]
&=\\frac{1}{6}n(n+1)(2n+7)-\\frac{1}{6}(n-1)\\cdot n\\cdot \\left\\{2(n-1)+7\\right\\}\\\\[1.5ex]
&=\\frac{n}{6}\\left\\{(n+1)(2n+7)-(n-1)(2n+5)\\right\\}\\\\[1.5ex]
&=\\frac{n}{6}\\left\\{2n^2+9n+7-(2n^2+3n-5)\\right\\}\\\\[1.5ex]
&=\\frac{n}{6}(6n+12)\\\\[1.5ex]
&=n(n+2)
\\end{align}
$$\u3068\u306a\u308b. \u3053\u308c\u306f, \\(n=1\\)\u306e\u3068\u304d\u3082\u6210\u308a\u7acb\u3064\u306e\u3067, \\(\\{a_n\\}\\)\u306e\u4e00\u822c\u9805\u306f$$
a_n=n(n+2)\\,\\,\\,(n=1,2,3,\\cdots)
$$\u3068\u6c42\u307e\u308b.<\/p>\n\n\n\n

(2) \u307e\u305a, \\(n=1\\)\u306e\u3068\u304d, $$
\\sum_{k=1}^1\\frac{1}{a_k}=\\frac{1}{a_1}=\\frac{1}{3}
$$\u3067\u3042\u308b. \u6b21\u306b, \\(n\\geq 2\\)\u306e\u3068\u304d,$$
\\begin{align}
\\sum_{k=1}^n\\frac{1}{a_k}&=\\sum_{k=1}^n\\frac{1}{k(k+2)}\\\\[1.5ex]
&=\\sum_{k=1}^n\\frac{1}{2}\\left(\\frac{1}{k}-\\frac{1}{k*2}\\right)\\\\[1.5ex]
&=\\frac{1}{2}\\left(1-\\frac{1}{3}\\right)+\\frac{1}{2}\\left(\\frac{1}{2}-\\frac{1}{4}\\right)+\\frac{1}{2}\\left(\\frac{1}{3}-\\frac{1}{5}\\right)+\\cdots\\\\[1.5ex]
&\\qquad +\\frac{1}{2}\\left(\\frac{1}{n-1}-\\frac{1}{n+1}\\right)+\\frac{1}{2}\\left(\\frac{1}{n}-\\frac{1}{n+2}\\right)\\\\[1.5ex]
&=\\frac{1}{2}\\left(1-\\cancel{\\frac{1}{3}}\\right)+\\frac{1}{2}\\left(\\frac{1}{2}-\\cancel{\\frac{1}{4}}\\right)+\\frac{1}{2}\\left(\\cancel{\\frac{1}{3}}-\\cancel{\\frac{1}{5}}\\right)+\\cdots\\\\[1.5ex]
&\\qquad +\\frac{1}{2}\\left(\\cancel{\\frac{1}{n-1}}-\\frac{1}{n+1}\\right)+\\frac{1}{2}\\left(\\cancel{\\frac{1}{n}}-\\frac{1}{n+2}\\right)\\\\[1.5ex]
&=\\frac{1}{2}\\left(1+\\frac{1}{2}-\\frac{1}{n+1}-\\frac{1}{n+2}\\right)\\\\[1.5ex]
&=\\frac{1}{2}\\cdot\\frac{3(n+1)(n+2)-2(n+2)-2(n+1)}{2(n+1)(n+2)}\\\\[1.5ex]
&=\\frac{3n^2+5n}{4(n+1)(n+2)}\\\\[1.5ex]
&=\\frac{n(3n+5)}{4(n+1)(n+2)}\\\\[1.5ex]
\\end{align}
$$\u3068\u306a\u308b. \u3053\u308c\u306f, \\(n=1\\)\u3067\u3082\u6210\u308a\u7acb\u3064\u306e\u3067, $$
\\sum_{k=1}^n\\frac{1}{a_k}=\\frac{n(3n+5)}{4(n+1)(n+2)}\\,\\,\\,(n=1,2,3,\\cdots)
$$\u3068\u6c42\u307e\u308b.<\/p>\n<\/div><\/div>\n\n\n\n

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