{"id":3817,"date":"2025-12-24T17:21:35","date_gmt":"2025-12-24T08:21:35","guid":{"rendered":"https:\/\/math-travel.com\/?p=3817"},"modified":"2026-03-13T12:35:19","modified_gmt":"2026-03-13T03:35:19","slug":"shigma","status":"publish","type":"post","link":"https:\/\/math-travel.jp\/math-b\/shigma\/","title":{"rendered":"\u03a3(\u30b7\u30b0\u30de)\u306e\u8a08\u7b97\u516c\u5f0f\u3068\u4f7f\u3044\u65b9\uff01\u6570\u5217\u306e\u548c\u306e\u6027\u8cea\u3068\u8a3c\u660e\u3092\u308f\u304b\u308a\u3084\u3059\u304f\u89e3\u8aac"},"content":{"rendered":"\n

\u6570\u5b66B\u306e\u6570\u5217\u3067\u51fa\u3066\u304f\u308b\u300c\u2211\uff08\u30b7\u30b0\u30de\uff09<\/span>\u300d\u306b\u60a9\u307e\u3055\u308c\u308b\u9ad8\u6821\u751f\u3082\u591a\u3044\u306f\u305a\u3002<\/span><\/p>\n\n\n\n

\u4eca\u56de\u89e3\u6c7a\u3059\u308b\u60a9\u307f<\/span><\/div>
\n

\u300c\u30b7\u30b0\u30de\u306e\u516c\u5f0f\u304c\u5206\u304b\u3089\u306a\u3044\u300d <\/p>\n\n\n\n

\u300c\u6570\u5217\u306e\u30b7\u30b0\u30de\u306e\u8a08\u7b97\u304c\u82e6\u624b\u300d<\/p>\n<\/div><\/div>\n\n\n\n

\u4eca\u56de\u306f\u6570\u5217\u306e\u30b7\u30b0\u30de\u306b\u95a2\u3059\u308b\u60a9\u307f\u3092\u89e3\u6c7a\u3057\u307e\u3059\u3002<\/p>\n\n\n

\"\"\u9ad8\u6821\u751f<\/span><\/div>
\n

\u03a3\u30b7\u30b0\u30de\u306e\u516c\u5f0f\u3092\u5fd8\u308c\u3066\u3057\u307e\u3063\u3066\u3001\u6570\u5217\u306e\u548c\u304c\u6c42\u3081\u3089\u308c\u306a\u3044\u2026<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n

\u6570\u5217\u306e\u548c\u3092\u6c42\u3081\u308b\u554f\u984c\u306a\u3069\u3001\u3055\u307e\u3056\u307e\u306a\u6240\u3067\u03a3\uff08\u30b7\u30b0\u30de\uff09<\/span>\u3092\u4f7f\u3044\u307e\u3059\u3002<\/p>\n\n\n\n

\u307e\u305a\u524d\u63d0\u306e\u77e5\u8b58\u3068\u3057\u3066\u3001\u03a3\uff08\u30b7\u30b0\u30de\uff09\u3068\u306f\u7dcf\u548c\u3092\u8868\u3059\u8a18\u53f7\u3067\u3001<\/p>\n\n\n\n

\\[\\displaystyle \\sum_{k=1}^{n} a_{k}=a_{1}+a_{2}+ \\cdots +a_{n}\\]<\/p>\n\n\n\n

\u3092\u8868\u3057\u3066\u3044\u307e\u3059\u3002<\/p>\n\n\n\n

\u4f8b\u3048\u3070\u3001\\(\\displaystyle \\sum_{k=3}^{10} a_{k}\\)\u306e\u3068\u304d\u306f\u3001\\(a_{n}\\)\u306en=3\u304b\u3089n=10\u307e\u3067\u306e\u8db3\u3057\u7b97\u3092\u610f\u5473\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

\\[\\displaystyle \\sum_{k=3}^{10} a_{k}=a_{3}+a_{4}+ \\cdots +a_{10}\\]<\/p>\n\n\n\n

\u305d\u3093\u306a\u30b7\u30b0\u30de\u306b\u306f\u7d76\u5bfe\u306b\u899a\u3048\u3066\u304a\u304d\u305f\u30445\u3064\u306e\u516c\u5f0f<\/span>\u304c\u3042\u308a\u307e\u3059\u3002<\/p>\n\n\n\n

\u03a3\u306e\u8a08\u7b97\u516c\u5f0f<\/span><\/div>
\n

\\(\\displaystyle 1. \\sum_{k=1}^{n} a=an\\)<\/p>\n\n\n\n

\\(\\displaystyle 2. \\sum_{k=1}^{n} k=\\frac{1}{2}n(n+1)\\)<\/p>\n\n\n\n

\\(\\displaystyle 3. \\sum_{k=1}^{n} k^{2}=\\frac{1}{6}n(n+1)(2n+1)\\)<\/p>\n\n\n\n

\\(\\displaystyle 4. \\sum_{k=1}^{n} k^{3}=\\{\\frac{1}{2}n(n+1)\\}^{2}\\)<\/p>\n\n\n\n

\\(\\displaystyle 5. \\sum_{k=1}^{n} ar^{k-1}=\\frac{a(r^{n}-1)}{r-1}=\\frac{a(1-r^{n})}{1-r}\\)<\/p>\n<\/div><\/div>\n\n\n\n

\u672c\u8a18\u4e8b\u3067\u306f\u03a3\u30b7\u30b0\u30de\u306e\u8a08\u7b97\u516c\u5f0f\u3068\u6027\u8cea\u306b\u3064\u3044\u3066\u89e3\u8aac<\/span>\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

\u03a3\u306e\u8a08\u7b97\u304c\u3067\u304d\u306a\u3044\u306e\u306f\u516c\u5f0f\u3092\u899a\u3048\u3066\u3044\u306a\u3044\u5834\u5408\u304c\u591a\u3044\u3067\u3059\u3002\u672c\u8a18\u4e8b\u3092\u8aad\u3093\u3067\u3001\u305c\u3072\u899a\u3048\u3066\u3057\u307e\u3044\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n

\u03a3\u30b7\u30b0\u30de\u306e\u8a08\u7b97\u516c\u5f0f<\/h2>\n\n\n\n

\u03a3\u30b7\u30b0\u30de\u3092\u5b66\u7fd2\u3059\u308b\u306b\u3042\u305f\u3063\u3066\u3001\u78ba\u5b9f\u306b\u899a\u3048\u3066\u304a\u304d\u305f\u3044\u516c\u5f0f\u304c5\u3064<\/span>\u3042\u308a\u307e\u3059\u3002<\/p>\n\n\n\n

\u03a3\u306e\u8a08\u7b97\u516c\u5f0f<\/span><\/div>
\n

\\(\\displaystyle 1. \\sum_{k=1}^{n} a=an\\)<\/p>\n\n\n\n

\\(\\displaystyle 2. \\sum_{k=1}^{n} k=\\frac{1}{2}n(n+1)\\)<\/p>\n\n\n\n

\\(\\displaystyle 3. \\sum_{k=1}^{n} k^{2}=\\frac{1}{6}n(n+1)(2n+1)\\)<\/p>\n\n\n\n

\\(\\displaystyle 4. \\sum_{k=1}^{n} k^{3}=\\{\\frac{1}{2}n(n+1)\\}^{2}\\)<\/p>\n\n\n\n

\\(\\displaystyle 5. \\sum_{k=1}^{n} ar^{k-1}=\\frac{a(r^{n}-1)}{r-1}=\\frac{a(1-r^{n})}{1-r}\\)<\/p>\n<\/div><\/div>\n\n\n\n

\u3069\u308c\u3082\u91cd\u8981\u306a\u516c\u5f0f\u306a\u306e\u3067\u3001\u5fc5\u305a\u899a\u3048\u307e\u3057\u3087\u3046\u3002<\/span><\/p>\n\n\n\n

\u30b7\u30b0\u30de\u306e\u8a08\u7b97\u516c\u5f0f\u306e\u8a3c\u660e\u306f\u300c4. \u03a3\u30b7\u30b0\u30de\u306e\u516c\u5f0f\u306e\u8a3c\u660e\u300d\u3067\u89e3\u8aac\u3057\u307e\u3059\u3002<\/p>\n\n\n

\"\"\u30b7\u30fc\u30bf<\/span><\/div>
\n

\u3053\u308c\u304b\u3089\u306f\u5f53\u305f\u308a\u524d\u306e\u3088\u3046\u306b\u516c\u5f0f\u3092\u4f7f\u3046\u304b\u3089\u306d<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n

\u03a3\u30b7\u30b0\u30de\u306e\u6027\u8cea<\/h2>\n\n\n\n

\u03a3\u30b7\u30b0\u30de\u306e\u8a08\u7b97\u516c\u5f0f\u3068\u5408\u308f\u305b\u3066\u3001\u4ee5\u4e0b\u306e\u6027\u8cea\u3082\u899a\u3048\u3066\u304a\u304d\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n

\u03a3\u30b7\u30b0\u30de\u306e\u6027\u8cea<\/span><\/div>
\n

\\(p,q\\)\u306f\u5b9a\u6570\u3068\u3059\u308b\u3068\u3001<\/p>\n\n\n\n

\\(\\displaystyle 1.\\sum_{k=1}^{n}(a_{k}+b_{k})=\\sum_{k=1}^{n} a_{k}+\\sum_{k=1}^{n} b_{k}\\)<\/p>\n\n\n\n

\\(\\displaystyle 2.\\sum_{k=1}^{n} pa_{k}=p\\sum_{k=1}^{n} a_{k}\\)<\/p>\n\n\n\n

1,2\u3088\u308a<\/p>\n\n\n\n

\\(\\displaystyle \\sum_{k=1}^{n}(pa_{k}+qb_{k})=p\\sum_{k=1}^{n} a_{k}+q\\sum_{k=1}^{n} b_{k}\\)<\/p>\n<\/div><\/div>\n\n\n\n

\u305d\u308c\u305e\u308c\u306e\u6027\u8cea\u306b\u3064\u3044\u3066\u8a3c\u660e\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

\u03a3\u30b7\u30b0\u30de\u306e\u6027\u8cea\u2460\u306e\u8a3c\u660e<\/span><\/div>
\n

\u6570\u5217\\(\\{a_{n}\\},\\{b_{n}\\}\\)\u306b\u5bfe\u3057\u3066\u3001<\/p>\n\n\n\n

\\(\\displaystyle \\sum_{k=1}^{n}(a_{k}+b_{k})\\)
\\(\\displaystyle =(a_{1}+b_{1})+(a_{2}+b_{2})+\\cdots+(a_{n}+b_{n})\\)
\\(\\displaystyle =(a_{1}+a_{2}+\\cdots+a_{n})+(b_{1}+b_{2}+\\cdots+b_{n})\\)
\\(\\displaystyle =\\sum_{k=1}^{n} a_{k}+\\sum_{k=1}^{n} b_{k}\\)<\/p>\n<\/div><\/div>\n\n\n\n

\u03a3\u30b7\u30b0\u30de\u306e\u6027\u8cea\u2461\u306e\u8a3c\u660e<\/span><\/div>
\n

\u6570\u5217\\(\\{a_{n}\\}\\)\u306b\u5bfe\u3057\u3066\u3001\\(p\\)\u3092\u5b9a\u6570\u3068\u3059\u308b\u3068<\/p>\n\n\n\n

\\(\\displaystyle \\sum_{k=1}^{n} pa_{k}\\)
\\(\\displaystyle =pa_{1}+pa_{2}+\\cdots+pa_{n}\\)
\\(\\displaystyle =p(a_{1}+a_{2}+\\cdots+a_{n})\\)
\\(\\displaystyle =p\\sum_{k=1}^{n} a_{k}\\)<\/p>\n<\/div><\/div>\n\n\n\n

\u03a3\u30b7\u30b0\u30de\u306e\u6027\u8cea\u2462\u306e\u8a3c\u660e<\/span><\/div>
\n

\u6570\u5217\\(\\{a_{n}\\},\\{b_{n}\\}\\)\u306b\u5bfe\u3057\u3066\u3001\\(p,q\\)\u3092\u5b9a\u6570\u3068\u3059\u308b\u3068<\/p>\n\n\n\n

\\(\\displaystyle \\sum_{k=1}^{n}(pa_{k}+qb_{k})\\)
\\(\\displaystyle =(pa_{1}+qb_{1})+(pa_{2}+qb_{2})+\\cdots+(pa_{n}+qb_{n})\\)
\\(\\displaystyle =p(a_{1}+a_{2}+\\cdots+a_{n})+q(b_{1}+b_{2}+\\cdots+b_{n})\\)
\\(\\displaystyle =p\\sum_{k=1}^{n} a_{k}+q\\sum_{k=1}^{n} b_{k}\\)<\/p>\n<\/div><\/div>\n\n\n\n

\u4ee5\u4e0a\u306e\u03a3\u306e\u8a08\u7b97\u516c\u5f0f\u3068\u6027\u8cea\u3092\u5229\u7528\u3059\u308c\u3070\u3001\u69d8\u3005\u306a\u6570\u5217\u306e\u548c\u3082\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u307e\u3059\u3002<\/p>\n\n\n

\"\"\u9ad8\u6821\u751f<\/span><\/div>
\n

\u304b\u3063\u3053\u306e\u4e2d\u8eab\u3092\u5206\u3051\u305f\u308a\u3067\u304d\u308b\u3093\u3060\u306d\uff01<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n

\"\"\u30b7\u30fc\u30bf<\/span><\/div>
\n

\u305d\u3046\u3044\u3046\u3053\u3068\uff01\u5de5\u592b\u3057\u3066\u8a08\u7b97\u3059\u308b\u306e\u304c\u5927\u4e8b\u3060\u3088\uff01<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n

\u03a3\u30b7\u30b0\u30de\u3092\u5229\u7528\u3059\u308b\u554f\u984c<\/h2>\n\n\n\n

\u03a3\u30b7\u30b0\u30de\u306e\u57fa\u672c\u554f\u984c<\/h3>\n\n\n\n

\u5b9f\u969b\u306b\u516c\u5f0f\u3084\u6027\u8cea\u3092\u4f7f\u3063\u3066\u3001\u3044\u304f\u3064\u304b\u554f\u984c\u3092\u89e3\u3044\u3066\u307f\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n

\u307e\u305a\u306f\u8d85\u57fa\u672c\u3068\u306a\u308b\u8a08\u7b97\u554f\u984c\u304b\u3089<\/p>\n\n\n\n

\u03a3\u30b7\u30b0\u30de\u306e\u57fa\u672c\u554f\u984c<\/span><\/div>
\n

\u6b21\u306e\u8a08\u7b97\u3092\u3057\u3066\u307f\u3088\u3046\u3002<\/p>\n\n\n\n

\\(\\displaystyle 1.\\sum_{k=1}^{n} 3k\\)<\/p>\n\n\n\n

\\(\\displaystyle 2.\\sum_{k=1}^{n} (k^{2}+2k)\\)<\/p>\n\n\n\n

\\(\\displaystyle 3.\\sum_{k=1}^{n} 3 \\cdot 2^{k}\\)<\/p>\n<\/div><\/div>\n\n\n\n

(1)\u306e\u89e3\u8aac<\/h4>\n\n\n\n

(1)\u306e\u554f\u984c\u306f\u5b9a\u6570\u306e3\u3092\u524d\u306b\u51fa\u3059\u3053\u3068\u3067\u3001\u03a3\u306e\u8a08\u7b97\u516c\u5f0f\u2461\u306e\u7b49\u5dee\u6570\u5217\u306e\u548c\u304c\u4f7f\u3048\u307e\u3059\u3002<\/p>\n\n\n\n

\\begin{eqnarray}
\\displaystyle \\sum_{k=1}^{n} 3k&=&3 \\sum_{k=1}^{n} k\\\\
\\displaystyle &=&3 \\cdot \\frac{1}{2}n(n+1)\\\\
\\displaystyle &=&\\frac{3}{2}n(n+1)
\\end{eqnarray}<\/p>\n\n\n\n

\u4f7f\u3063\u305f\u516c\u5f0f<\/span><\/div>
\n

\\[\\displaystyle \\sum_{k=1}^{n} k=\\frac{1}{2}n(n+1)\\]<\/p>\n<\/div><\/div>\n\n\n\n

(2)\u306e\u89e3\u8aac<\/h4>\n\n\n\n

(2)\u306e\u554f\u984c\u306f\u304b\u3063\u3053\u306e\u306a\u304b\u3092\u5206\u3051\u3066\u8a08\u7b97\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

\\begin{eqnarray}
\\displaystyle \\sum_{k=1}^{n} (k^{2}+2k)&=&\\sum_{k=1}^{n} k^{2}+\\sum_{k=1}^{n} 2k\\\\
\\displaystyle &=&\\sum_{k=1}^{n} k^{2}+2\\sum_{k=1}^{n} k\\\\
\\displaystyle &=&\\frac{1}{6}n(n+1)(2n+1)+2 \\cdot \\frac{1}{2}n(n+1)\\\\
\\displaystyle &=&\\frac{1}{6}n(n+1)(2n+7)
\\end{eqnarray}<\/p>\n\n\n\n

\u4f7f\u3063\u305f\u516c\u5f0f<\/span><\/div>
\n

\\[\\displaystyle \\sum_{k=1}^{n} k=\\frac{1}{2}n(n+1)\\]<\/p>\n\n\n\n

\\[\\displaystyle \\sum_{k=1}^{n} k^{2}=\\frac{1}{6}n(n+1)(2n+1)\\]<\/p>\n<\/div><\/div>\n\n\n\n

(3)\u306e\u89e3\u8aac<\/h4>\n\n\n\n

(3)\u306e\u554f\u984c\u306f\u8a08\u7b97\u516c\u5f0f\u2464\u3067\u7d39\u4ecb\u3057\u305f\u7b49\u6bd4\u6570\u5217\u306e\u548c\u3068\u3057\u3066\u8003\u3048\u307e\u3059\u3002<\/p>\n\n\n\n

\\begin{eqnarray}
\\displaystyle \\sum_{k=1}^{n} 3 \\cdot 2^{k}&=&3 \\sum_{k=1}^{n} 2^{k}\\\\
\\displaystyle &=&3 \\cdot \\frac{2(2^{n}-1)}{2-1}\\\\
\\displaystyle &=&6(2^{n}-1)
\\end{eqnarray}<\/p>\n\n\n\n

\u7b49\u5dee\u6570\u5217\u3084\u7b49\u6bd4\u6570\u5217\u306e\u548c\u306e\u516c\u5f0f\u3092\u307e\u3060\u899a\u3048\u3066\u306a\u3044\u4eba\u306f\u3059\u3050\u306b\u78ba\u8a8d\u3057\u307e\u3057\u3087\u3046\uff01<\/p>\n\n\n

\"\"\u30b7\u30fc\u30bf<\/span><\/div>
\n

\u3053\u308c\u306f\u516c\u5f0f\u3092\u899a\u3048\u3066\u30b9\u30e9\u30b9\u30e9\u3068\u89e3\u3051\u3066\u6b32\u3057\u3044\u306a<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n

\"\"\u9ad8\u6821\u751f<\/span><\/div>
\n

\u516c\u5f0f\u3092\u899a\u3048\u305f\u304b\u3089\u8a08\u7b97\u306a\u3089\u3067\u304d\u305d\u3046\uff01<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n

\u4e00\u822c\u9805\u3092\u6c42\u3081\u3066\u304b\u3089\u548c\u3092\u6c42\u3081\u308b\u554f\u984c<\/h3>\n\n\n\n

\u4ee5\u4e0b\u306e\u3088\u3046\u306a\u4e00\u822c\u9805\u3092\u6c42\u3081\u3066\u304b\u3089\u548c\u3092\u6c42\u3081\u308b\u554f\u984c\u3082\u3042\u308a\u307e\u3059\u3002<\/p>\n\n\n\n

\u6570\u5217\u306e\u548c<\/span><\/div>
\n

\u6b21\u306e\u6570\u5217\u306e\u521d\u9805\u304b\u3089\u7b2c\\(n\\)\u9805\u307e\u3067\u306e\u548c\u3092\u6c42\u3081\u3088\u3046\u3002<\/p>\n\n\n\n

\\[1 , 6 , 11 , 16 , 21 , \u2026\\]<\/p>\n<\/div><\/div>\n\n\n\n

\u3053\u306e\u554f\u984c\u306f\u307e\u305a\u6570\u5217\u306e\u4e00\u822c\u9805\\(a_{n}\\)\u3092\u6c42\u3081\u308b\u5fc5\u8981\u304c\u3042\u308a\u307e\u3059\u3002<\/span><\/p>\n\n\n\n

\u306a\u305c\u306a\u3089\u3001\u4e00\u822c\u9805\u3067\u8868\u3059\u3053\u3068\u3067\u03a3\u306e\u8a08\u7b97\u304c\u3067\u304d\u308b\u3088\u3046\u306b\u306a\u308b\u304b\u3089\u3067\u3059\u3002<\/p>\n\n\n\n

1 , 6 , 11 , 16 , 21 , \u2026<\/p>\n\n\n\n

\u4e0e\u3048\u3089\u308c\u305f\u6570\u5217\u306f\u300c\u521d\u98051\u3001\u516c\u5dee5\u306e\u7b49\u5dee\u6570\u5217\u300d\u3067\u3059\u3002<\/p>\n\n\n\n

\u3057\u305f\u304c\u3063\u3066\u3001\u4e00\u822c\u9805\\(a_{n}\\)\u306f<\/p>\n\n\n\n

\\begin{eqnarray}
a_{n}&=&1+5(n-1)\\\\
&=&5n-4
\\end{eqnarray}<\/p>\n\n\n\n

\u4e00\u822c\u9805\u304c\u5206\u304b\u308c\u3070\u3042\u3068\u306f\u03a3\u306e\u8a08\u7b97\u3092\u3059\u308b\u3060\u3051\u3067\u3059\u3002<\/p>\n\n\n\n

\\begin{eqnarray}
\\displaystyle \\sum_{k=1}^{n} a_{k}&=&\\sum_{k=1}^{n} (5k-4)\\\\
\\displaystyle &=&5 \\sum_{k=1}^{n} k – \\sum_{k=1}^{n} 4\\\\
\\displaystyle &=&5 \\cdot \\frac{1}{2}n(n+1) – 4n\\\\
\\displaystyle &=&\\frac{1}{2}n(5n-3)
\\end{eqnarray}<\/p>\n\n\n\n

\u3057\u305f\u304c\u3063\u3066\u3001\u4e0e\u3048\u3089\u308c\u305f\u6570\u5217\u306e\u521d\u9805\u304b\u3089\u7b2c\\(n\\)\u9805\u307e\u3067\u306e\u548c\u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u307e\u3057\u305f\u3002<\/p>\n\n\n\n

\u6570\u5217\u306e\u4e00\u822c\u9805\u3092\u6c42\u3081\u308b<\/h3>\n\n\n\n

\u6570\u5217\u306e\u4e00\u822c\u9805\u3092\u6c42\u3081\u308b\u3068\u304d\u306b\u3082\u03a3\u3092\u4f7f\u3044\u307e\u3059\u3002<\/p>\n\n\n

\n
\"\u968e\u5dee\u6570\u5217\"<\/figure>\n<\/div>\n\n\n

\u6570\u5217\\(\\{a_{n}\\}\\)\u306e\u96a3\u308a\u5408\u3046\u9805\u306e\u5dee\u3067\u3067\u304d\u305f\u6570\u5217\\(\\{b_{n}\\}\\)\u3092\u968e\u5dee\u6570\u5217\u3068\u3044\u3044\u307e\u3059\u3002<\/p>\n\n\n\n

\u3082\u3068\u306e\u6570\u5217\u306e\u4e00\u822c\u9805\\(a_{n}\\)\u3092\u6c42\u3081\u308b\u306b\u306f\u3001\u521d\u9805\\(a_{1}\\)\u306b\u5bfe\u3057\u3066\u968e\u5dee\u6570\u5217\\(\\{b_{n}\\}\\)\u3092\u52a0\u3048\u308b\u30a4\u30e1\u30fc\u30b8\u3067\u8003\u3048\u307e\u3059\u3002<\/p>\n\n\n\n

\n

\\(a_{1}=a_{1}\\)\u3000\u2190\u521d\u9805<\/p>\n\n\n\n

\\(a_{2}=a_{1}+b_{1}\\)<\/p>\n\n\n\n

\\(\\displaystyle a_{3}=a_{2}+b_{2}=a_{1}+\\sum_{k=1}^{2} b_{k}\\)<\/p>\n\n\n\n

\\(\\displaystyle a_{4}=a_{3}+b_{3}=a_{1}+\\sum_{k=1}^{3} b_{k}\\)<\/p>\n\n\n\n

\uff1a<\/p>\n\n\n\n

\\(\\displaystyle a_{n}=a_{n-1}+b_{n-1}=a_{1}+\\sum_{k=1}^{n-1} b_{k}\\)<\/p>\n<\/div><\/div>\n\n\n\n

\u3057\u305f\u304c\u3063\u3066\u3001\u5143\u306e\u6570\u5217\u306e\u4e00\u822c\u9805\\(a_{n}\\)\u306f<\/p>\n\n\n\n

\\[\\displaystyle a_{n}=a_{n-1}+b_{n-1}=a_{1}+\\sum_{k=1}^{n-1} b_{k}\\]<\/p>\n\n\n\n

\u3067\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u307e\u3059\u3002<\/p>\n\n\n\n

\u968e\u5dee\u6570\u5217\u306b\u3064\u3044\u3066\u306f\u300c\u968e\u5dee\u6570\u5217\u3092\u7528\u3044\u305f\u4e00\u822c\u9805\u3068\u548c\u3092\u6c42\u3081\u308b\u516c\u5f0f\uff01\u300d\u3067\u8a73\u3057\u304f\u307e\u3068\u3081\u307e\u3057\u305f\u3002<\/p>\n\n\n\n

\u5206\u6570\u306e\u548c\u306f\u90e8\u5206\u5206\u6570\u5206\u89e3<\/h3>\n\n\n\n

\u4ee5\u4e0b\u306e\u3088\u3046\u306a\u5206\u6570\u306e\u548c\u306f\u30b7\u30b0\u30de\u3067\u306f\u306a\u304f\u90e8\u5206\u5206\u6570\u5206\u89e3<\/span>\u3067\u89e3\u304d\u307e\u3059\u3002<\/p>\n\n\n\n

\\[\\displaystyle S=\\frac{1}{1 \\cdot 2}+\\frac{1}{2 \\cdot 3}+\\frac{1}{3 \\cdot 4}+\\cdots+\\frac{1}{n(n+1)}\\]<\/p>\n\n\n

\"\"\u9ad8\u6821\u751f<\/span><\/div>
\n

\u90e8\u5206\u5206\u6570\u5206\u89e3\u2026\u6f22\u5b57\u304c\u591a\u304f\u3066\u96e3\u3057\u305d\u3046\u2026<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n

\u306a\u3093\u3060\u304b\u96e3\u3057\u305d\u3046\u3068\u601d\u3063\u305f\u65b9\u3082\u591a\u3044\u3067\u3059\u3088\u306d\u3002<\/p>\n\n\n\n

\u90e8\u5206\u5206\u6570\u5206\u89e3\u3068\u306f\u3001\u4ee5\u4e0b\u306e\u3088\u3046\u306b1\u3064\u306e\u5206\u6570\u3092\u5206\u89e3\u3059\u308b\u3053\u3068\u3092\u6307\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

\u90e8\u5206\u5206\u6570\u5206\u89e3<\/span><\/div>
\n

\\[\\displaystyle \\frac{1}{k(k+1)}=\\frac{1}{k}-\\frac{1}{k+1}\\]<\/p>\n<\/div><\/div>\n\n\n\n

\u90e8\u5206\u5206\u6570\u5206\u89e3\u3092\u5229\u7528\u3059\u308b\u3068\u3001\u4e0e\u3048\u3089\u308c\u305f\u5206\u6570\u306e\u548c\u3082<\/p>\n\n\n\n

\n

\\begin{eqnarray}
\\displaystyle S&=&\\frac{1}{1 \\cdot 2}+\\frac{1}{2 \\cdot 3}+\\frac{1}{3 \\cdot 4}+\\cdots+\\frac{1}{n(n+1)}\\\\
\\displaystyle &=&(\\frac{1}{1}-\\frac{1}{2})+(\\frac{1}{2}-\\frac{1}{3})+(\\frac{1}{3}-\\frac{1}{4})+\\cdots+(\\frac{1}{n}-\\frac{1}{n+1})\\\\
\\displaystyle &=&1-\\frac{1}{n+1}\\\\
\\displaystyle &=&\\frac{n}{n+1}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n

\u305d\u308c\u305e\u308c\u306e\u5206\u6570\u304c\u7b26\u53f7\u306e\u9055\u3046\u3082\u306e\u540c\u58eb\u3067\u76f8\u6bba\u3059\u308b\u306e\u3067\u3001\u30b7\u30f3\u30d7\u30eb\u306a\u7b54\u3048\u306b\u306a\u308a\u307e\u3059\u3002<\/p>\n\n\n

\"\"\u9ad8\u6821\u751f<\/span><\/div>
\n

\u3053\u308c\u306f\u90e8\u5206\u5206\u6570\u5206\u89e3\u3092\u77e5\u3089\u306a\u3044\u3068\u56f0\u3063\u3061\u3083\u3046\u306d<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n

\"\"\u30b7\u30fc\u30bf<\/span><\/div>
\n

\u305d\u3046\u306a\u3093\u3060\u3088\u3002\u5206\u6570\u306e\u548c\u3092\u307f\u305f\u3089\u90e8\u5206\u5206\u6570\u5206\u89e3\u3092\u7591\u3063\u3066\u307f\u3088\u3046<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n

4. \u03a3\u30b7\u30b0\u30de\u306e\u516c\u5f0f\u306e\u8a3c\u660e<\/h2>\n\n\n\n

\u300c1.\u03a3\u30b7\u30b0\u30de\u306e\u8a08\u7b97\u516c\u5f0f<\/a>\u300d\u3067\u7d39\u4ecb\u3057\u305f\u03a3\u30b7\u30b0\u30de\u306e\u516c\u5f0f\u3092\u8a3c\u660e\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

\u8a3c\u660e\u3092\u8aad\u307e\u306a\u3044\u65b9\u306f\u98db\u3070\u3057\u3066\u3082\u3089\u3063\u3066\u5927\u4e08\u592b\u306a\u3068\u3053\u308d\u3067\u3059\u3002<\/p>\n\n\n\n

\u21d2\u8a3c\u660e\u3092\u98db\u3070\u3059<\/a><\/p>\n\n\n\n

\u03a3\u30b7\u30b0\u30de\u306e\u8a08\u7b97\u516c\u5f0f<\/span><\/div>
\n

\\(\\displaystyle 1. \\sum_{k=1}^{n} a=an\\)<\/p>\n\n\n\n

\\(\\displaystyle 2. \\sum_{k=1}^{n} k=\\frac{1}{2}n(n+1)\\)<\/p>\n\n\n\n

\\(\\displaystyle 3. \\sum_{k=1}^{n} k^{2}=\\frac{1}{6}n(n+1)(2n+1)\\)<\/p>\n\n\n\n

\\(\\displaystyle 4. \\sum_{k=1}^{n} k^{3}=\\{\\frac{1}{2}n(n+1)\\}^{2}\\)<\/p>\n\n\n\n

\\(\\displaystyle 5. \\sum_{k=1}^{n} ar^{k-1}=\\frac{a(r^{n}-1)}{r-1}=\\frac{a(1-r^{n})}{1-r}\\)<\/p>\n<\/div><\/div>\n\n\n\n

\u516c\u5f0f\u306e\u8a3c\u660e\u2460<\/span><\/div>
\n

\\(a\\)\u304c\\(k\\)\u3092\u542b\u3093\u3067\u3044\u306a\u3044\u306e\u3067\\(\\displaystyle \\sum_{k=1}^{n} a\\)\u3068\u3044\u3046\u306e\u306f\u3001\\(a\\)\u3092\\(n\\)\u56de\u8db3\u3059\u3053\u3068\u3092\u610f\u5473\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

\u3057\u305f\u304c\u3063\u3066\u3001<\/p>\n\n\n\n

\\begin{eqnarray}
\\displaystyle \\sum_{k=1}^{n} a&=&a+a+ \\cdot + a\\\\
&=&an
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n

\u516c\u5f0f\u306e\u8a3c\u660e\u2461<\/span><\/div>
\n

\\[\\displaystyle \\sum_{k=1}^{n} k=1+2+3+\\cdots +n\\]<\/p>\n\n\n\n

\u3053\u308c\u306f\u300c\u521d\u98051\u3001\u672b\u9805\\(n\\)\u3001\u9805\u6570\\(n\\)\u306e\u7b49\u5dee\u6570\u5217\u306e\u548c\u300d\u3092\u8868\u3057\u3066\u3044\u307e\u3059\u3002<\/p>\n\n\n\n

\u53f3\u8fba\u3092\u7b49\u5dee\u6570\u5217\u306e\u548c\u306e\u516c\u5f0f\u306b\u7f6e\u304d\u63db\u3048\u3066<\/p>\n\n\n\n

\\begin{eqnarray}
\\displaystyle \\sum_{k=1}^{n} k&=&1+3+ \\cdot + n\\\\
\\displaystyle &=&\\frac{1}{2}n(n+1)
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n

\u516c\u5f0f\u306e\u8a3c\u660e\u2462<\/span><\/div>
\n

\u6052\u7b49\u5f0f<\/p>\n\n\n\n

\\[(k+1)^{3}-k^{3}=3k^{3}+3k+1\\]<\/p>\n\n\n\n

\u3092\u5229\u7528\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

\\(k=1,2,3,\\cdots,n\\)\u3068\u3059\u308b\u3068<\/p>\n\n\n\n

\"\u30b7\u30b0\u30de\u306e\u516c\u5f0f\u8a3c\u660e\u2460\"<\/p>\n\n\n\n

\u3053\u308c\u3089\u306e\u5f0f\u306e\u4e21\u8fba\u3092\u52a0\u3048\u308b\u3068\u3001<\/p>\n\n\n\n

\"\u30b7\u30b0\u30de\u306e\u516c\u5f0f\u8a3c\u660e\u2461\"<\/p>\n\n\n\n

\u3053\u308c\u3089\u3092\u79fb\u9805\u3057\u3066\u3001<\/p>\n\n\n\n

\\[\\displaystyle 3\\sum_{k=1}^{n} k^{2}=(n+1)^{3}-1-3\\sum_{k=1}^{n} k -n\\]<\/p>\n\n\n\n

\u3053\u3053\u3067\\(\\displaystyle \\sum_{k=1}^{n} k =\\frac{1}{2}n(n+1)\\)\u3092\u4ee3\u5165\u3057\u3066
\\begin{eqnarray}
\\displaystyle 3 \\sum_{k=1}^{n} k^{2} &=& (n+1)^{3}-1-3 \\cdot \\frac{1}{2}n(n+1) -n\\\\
\\displaystyle &=& n^{3}+3 n^{2}+3 n-\\frac{3}{2} n^{2}-\\frac{5}{2} n \\\\
\\displaystyle &=& n^{3}+\\frac{3}{2} n^{2}+\\frac{1}{2} n \\\\
\\displaystyle &=& \\frac{1}{2} n\\left(2 n^{2}+3 n+1\\right) \\\\
\\displaystyle &=& \\frac{1}{2} n(n+1)(2 n+1)
\\end{eqnarray}
\u6700\u5f8c\u306b\u4e21\u8fba\u30923\u3067\u5272\u3063\u3066\u3001<\/p>\n\n\n\n

\\[\\displaystyle  \\sum_{k=1}^{n} k^{2}=\\frac{1}{6}n(n+1)(2n+1)\\]<\/p>\n<\/div><\/div>\n\n\n\n

\u516c\u5f0f\u306e\u8a3c\u660e\u2463<\/span><\/div>
\n

\u6052\u7b49\u5f0f<\/p>\n\n\n\n

\\[(k+1)^{4}-k^{4}=4k^{3}+6k^{2}+4k+1\\]<\/p>\n\n\n\n

\u3092\u5229\u7528\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

\\(k=1,2,3,\\cdots,n\\)\u3068\u3059\u308b\u3068<\/p>\n\n\n

\n
\"\u30b7\u30b0\u30de\u306e\u516c\u5f0f\u8a3c\u660e\u2462\"<\/figure>\n<\/div>\n\n\n

\u3053\u308c\u3089\u306e\u5f0f\u306e\u4e21\u8fba\u3092\u52a0\u3048\u308b\u3068\u3001<\/p>\n\n\n\n

\"\u30b7\u30b0\u30de\u306e\u516c\u5f0f\u8a3c\u660e\u2463\"<\/p>\n\n\n\n

\u3053\u308c\u3089\u3092\u79fb\u9805\u3057\u3066\u3001
\\[\\displaystyle 4\\sum_{k=1}^{n} k^{3}=(n+1)^{4}-1-6\\sum_{k=1}^{n} k^{2} -4\\sum_{k=1}^{n} k -n\\]
\u3053\u3053\u3067<\/p>\n\n\n\n

\\(\\displaystyle \\sum_{k=1}^{n} k =\\frac{1}{2}n(n+1)\\)
\\(\\displaystyle  \\sum_{k=1}^{n} k^{2}=\\frac{1}{6}n(n+1)(2n+1)\\)<\/p>\n\n\n\n

\u3092\u4ee3\u5165\u3057\u3066\u3001\u5f0f\u3092\u6574\u7406\u3059\u308b\u3068\u3001<\/p>\n\n\n\n

\\[\\displaystyle 4\\sum_{k=1}^{n} k^{3}=n^{2}(n+1)^{2}\\]<\/p>\n\n\n\n

\u6700\u5f8c\u306b\u4e21\u8fba\u30924\u3067\u5272\u3063\u3066\u3001<\/p>\n\n\n\n

\\[\\displaystyle \\sum_{k=1}^{n} k^{3}=\\{\\frac{1}{2}n(n+1)\\}^{2}\\]<\/p>\n<\/div><\/div>\n\n\n\n

\u516c\u5f0f\u306e\u8a3c\u660e\u2464<\/span><\/div>
\n

\\[\\displaystyle \\sum_{k=1}^{n} ar^{k-1}=a+ar+\\cdots +ar^{n-1}\\]<\/p>\n\n\n\n

\u3053\u308c\u306f\u300c\u521d\u9805\\(a\\)\u3001\u516c\u6bd4\\(r\\)\u3001\u9805\u6570\\(n-1\\)\u306e\u7b49\u6bd4\u6570\u5217\u306e\u548c\u300d\u3092\u8868\u3057\u3066\u3044\u307e\u3059\u3002<\/p>\n\n\n\n

\u53f3\u8fba\u3092\u7b49\u6bd4\u6570\u5217\u306e\u548c\u306e\u516c\u5f0f\u306b\u7f6e\u304d\u63db\u3048\u308b\u3068<\/p>\n\n\n\n

\\begin{eqnarray}
\\displaystyle \\sum_{k=1}^{n} ar^{k-1}&=&a+ar+\\cdots +ar^{n-1}\\\\
\\displaystyle &=&\\frac{a(r^{n}-1)}{r-1}\u3000(r > 0)\\\\
\\displaystyle &=&\\frac{a(1-r^{n})}{1-r}\u3000(r < 0)
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n

\u516c\u5f0f\u306e\u8a3c\u660e\u306f\u3067\u304d\u306a\u304f\u3066\u3082\u554f\u984c\u306a\u3044\u306e\u3067\u3001\u307e\u305a\u306f5\u3064\u306e\u516c\u5f0f\u3092\u78ba\u5b9f\u306b\u4f7f\u3048\u308b\u3088\u3046\u306b\u306a\u308a\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n

\u300a\u91cd\u8981\u300b3\u3064\u306e\u57fa\u672c\u6570\u5217<\/h2>\n\n\n\n

\u4eca\u56de\u306f\u03a3\u30b7\u30b0\u30de\u306e\u516c\u5f0f\u3084\u6027\u8cea\u306b\u3064\u3044\u3066\u89e3\u8aac\u3057\u307e\u3057\u305f\u3002<\/span><\/p>\n\n\n\n

\u30b7\u30b0\u30de\u306f\u6570\u5217\u306e\u548c\u3092\u6c42\u3081\u308b\u3068\u304d\u306b\u6d3b\u8e8d\u3057\u307e\u3059\u304c\u3001\u305d\u3082\u305d\u3082\u6570\u5217\u306e\u4e00\u822c\u9805\u3092\u6c42\u3081\u3089\u308c\u306a\u3044\u3068\u30b7\u30b0\u30de\u3092\u6d3b\u7528\u3067\u304d\u307e\u305b\u3093\u3002<\/p>\n\n\n\n

\u4ee5\u4e0b\u306e\u6570\u5217\u306f\u304b\u306a\u308a\u91cd\u8981\u306a\u306e\u3067\u78ba\u5b9f\u306b\u62bc\u3055\u3048\u3066\u304a\u304d\u307e\u3057\u3087\u3046\u3002<\/span><\/p>\n\n\n\n

\u57fa\u672c\u306e\u6570\u5217<\/span><\/div>
\n
    \n
  • \u7b49\u5dee\u6570\u5217<\/li>\n\n\n\n
  • \u7b49\u6bd4\u6570\u5217<\/li>\n\n\n\n
  • \u968e\u5dee\u6570\u5217<\/li>\n<\/ul>\n<\/div><\/div>\n\n\n\n

    \u30fb\u7b49\u5dee\u6570\u5217<\/span><\/p>\n\n\n\n

    \u7b49\u5dee\u6570\u5217\u3068\u306f\u3001\u300c\u4e00\u5b9a\u306e\u5dee\u3067\u5909\u5316\u3059\u308b\u6570\u5217\u300d<\/span>\u3092\u6307\u3057\u307e\u3059\u3002<\/p>\n\n\n

    \n
    \"\u7b49\u5dee\u6570\u5217\"<\/figure>\n<\/div>\n\n\n

    \u30fb\u7b49\u6bd4\u6570\u5217<\/span><\/p>\n\n\n\n

    \u7b49\u6bd4\u6570\u5217\u3068\u306f\u3001\u300c\u521d\u3081\u306e\u9805\u306b\u4e00\u5b9a\u306e\u6570\u3092\u304b\u3051\u7d9a\u3051\u3066\u3044\u304f\u6570\u5217\u300d<\/span>\u3092\u6307\u3057\u307e\u3059\u3002<\/p>\n\n\n

    \n
    \"\u7b49\u6bd4\u6570\u5217\"<\/figure>\n<\/div>\n\n\n

    \u30fb\u968e\u5dee\u6570\u5217<\/span><\/p>\n\n\n\n

    \u968e\u5dee\u6570\u5217\u306f\u8907\u96d1\u3067\u3001\u5404\u9805\u306e\u5dee\u3092\u66f8\u304d\u51fa\u3057\u3066\u307f\u308b\u3068\u3042\u308b\u6570\u5217\u304c\u898b\u3048\u3066\u304d\u307e\u3059\u3002<\/p>\n\n\n

    \n
    \"\u968e\u5dee\u6570\u5217\"<\/figure>\n<\/div>\n\n\n

    \u4e0a\u306e\u6570\u5217\u306e\u5834\u5408\u3001\u5404\u9805\u306e\u5dee\u304c\u7b49\u5dee\u6570\u5217\u306b\u306a\u3063\u3066\u3044\u307e\u3059\u3002<\/span><\/p>\n\n\n\n

    \u3053\u306e\u5dee\u304c\u7b49\u6bd4\u6570\u5217\u306b\u306a\u308b\u5834\u5408\u3082\u3042\u308a\u307e\u3059\u3057\u3001\u3082\u3063\u3068\u8907\u96d1\u306a\u6570\u5217\u306b\u306a\u308b\u3068\u304d\u3082\u3042\u308a\u307e\u3059\u3002<\/p>\n\n\n\n

    \u03a3\u30b7\u30b0\u30de\u306e\u516c\u5f0f\u3000\u307e\u3068\u3081<\/h2>\n\n\n\n

    \u4eca\u56de\u306f\u03a3\u30b7\u30b0\u30de\u306e\u8a08\u7b97\u516c\u5f0f\u3084\u6027\u8cea\u306b\u3064\u3044\u3066\u307e\u3068\u3081\u307e\u3057\u305f\u3002<\/span><\/p>\n\n\n\n

    \u03a3\u306e\u8a08\u7b97\u516c\u5f0f<\/span><\/div>
    \n

    \\(\\displaystyle 1. \\sum_{k=1}^{n} a=an\\)<\/p>\n\n\n\n

    \\(\\displaystyle 2. \\sum_{k=1}^{n} k=\\frac{1}{2}n(n+1)\\)<\/p>\n\n\n\n

    \\(\\displaystyle 3. \\sum_{k=1}^{n} k^{2}=\\frac{1}{6}n(n+1)(2n+1)\\)<\/p>\n\n\n\n

    \\(\\displaystyle 4. \\sum_{k=1}^{n} k^{3}=\\{\\frac{1}{2}n(n+1)\\}^{2}\\)<\/p>\n\n\n\n

    \\(\\displaystyle 5. \\sum_{k=1}^{n} ar^{k-1}=\\frac{a(r^{n}-1)}{r-1}=\\frac{a(1-r^{n})}{1-r}\\)<\/p>\n<\/div><\/div>\n\n\n\n

    \u03a3\u30b7\u30b0\u30de\u306e\u6027\u8cea<\/span><\/div>
    \n

    \\(p,q\\)\u306f\u5b9a\u6570\u3068\u3059\u308b\u3068\u3001<\/p>\n\n\n\n

    \\(\\displaystyle 1.\\sum_{k=1}^{n}(a_{k}+b_{k})=\\sum_{k=1}^{n} a_{k}+\\sum_{k=1}^{n} b_{k}\\)<\/p>\n\n\n\n

    \\(\\displaystyle 2.\\sum_{k=1}^{n} pa_{k}=p\\sum_{k=1}^{n} a_{k}\\)<\/p>\n\n\n\n

    1,2\u3088\u308a<\/p>\n\n\n\n

    \\(\\displaystyle \\sum_{k=1}^{n}(pa_{k}+qb_{k})=p\\sum_{k=1}^{n} a_{k}+q\\sum_{k=1}^{n} b_{k}\\)<\/p>\n<\/div><\/div>\n\n\n\n

    \u6570\u5217\u306e\u5358\u5143\u306f\u899a\u3048\u308b\u3053\u3068\u306f\u591a\u3044\u3067\u3059\u304c\u3001\u554f\u984c\u306e\u30d1\u30bf\u30fc\u30f3\u304c\u9650\u3089\u308c\u3066\u3044\u307e\u3059\u3002<\/p>\n\n\n\n

    \u305d\u308c\u305e\u308c\u306e\u6027\u8cea\u3084\u516c\u5f0f\u3092\u3057\u3063\u304b\u308a\u3068\u899a\u3048\u308c\u3070\u3001\u6570\u5217\u306f\u30d9\u30af\u30c8\u30eb\u3088\u308a\u3082\u5f97\u70b9\u3057\u3084\u3059\u3044\u5358\u5143\u3067\u3059\u3002<\/span><\/p>\n\n\n

    \"\"\u9ad8\u6821\u751f<\/span><\/div>
    \n

    \u03a3\u306e\u8a08\u7b97\u304c\u82e6\u624b\u3060\u3068\u601d\u3063\u3066\u3044\u305f\u3051\u3069\u3001\u516c\u5f0f\u3092\u899a\u3048\u3066\u3044\u306a\u3044\u3060\u3051\u3060\u3063\u305f\u3093\u3060\u306d\uff01<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n

    \"\"\u30b7\u30fc\u30bf<\/span><\/div>
    \n

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