\u6570\u5217 \\(\\{ a _ k \\}\\) \u3092 \\(a _ k = k +\\cos \\left( \\dfrac{k \\pi}{6} \\right)\\) \u3067\u5b9a\u3081\u308b. \\(n\\) \u3092\u6b63\u306e\u6574\u6570\u3068\u3059\u308b.<\/p>\r\n
(1)<\/strong>\u3000\\(\\textstyle\\sum\\limits _ {k=1}^{12n} a _ k\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(\\textstyle\\sum\\limits _ {k=1}^{12n} {a _ k}^2\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\(\\cos ( x +\\pi ) = \\cos x\\) , \\(\\cos \\left( x -\\dfrac{\\pi}{2} \\right) = -\\cos x\\) \u3092\u7528\u3044\u308c\u3070\r\n\\[\r\n\\textstyle\\sum\\limits _ {k=1}^{12n} \\cos \\left( \\dfrac{k \\pi}{6} \\right) = n ( -1 +1 ) = 0\r\n\\]\r\n\u3088\u3063\u3066\r\n\\[\\begin{align}\r\n\\textstyle\\sum\\limits _ {k=1}^{12n} a _ k & = \\textstyle\\sum\\limits _ {k=1}^{12n} k \\\\\r\n& = \\underline{6n ( 12n+1 )}\r\n\\end{align}\\]\r\n (2)<\/strong><\/p>\r\n \\[\\begin{align}\r\n{a _ k}^2 & = k^2 +2 k \\cos \\dfrac{k \\pi}{6} +\\cos^2 \\dfrac{k \\pi}{6} \\\\\r\n& = k^2 +2 \\underline{k \\cos \\dfrac{k \\pi}{6}} _ {[1]} +\\dfrac{1}{2} \\underline{\\cos \\dfrac{k \\pi}{3}} _ {[2]} +\\dfrac{1}{2}\r\n\\end{align}\\]\r\n\u3053\u3053\u3067, \u4e0b\u7dda\u90e8 [1] \u306b\u3064\u3044\u3066, \u6574\u6570 \\(m\\) \u306b\u5bfe\u3057\u3066\r\n\\[\\begin{align}\r\n\\textstyle\\sum\\limits _ {k = 12m+1}^{12(m+1)} [1] & = \\dfrac{\\sqrt{3}}{2} \\left\\{ (12m+1) +(12m+11) -(12m+5) -(12m+7) \\right\\} \\\\\r\n& \\quad +\\dfrac{1}{2} \\left\\{ (12m+2) +(12m+10) -(12m+4) -(12m+8) \\right\\} \\\\ & \\qquad -1 \\cdot (12m+6) +1 \\cdot (12m+12) \\\\\r\n& = 6\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\r\n\\textstyle\\sum\\limits _ {k=1}^{12n} [1] = 6n\r\n\\]\r\n\u307e\u305f, \u4e0b\u7dda\u90e8 [2] \u306b\u3064\u3044\u3066, (1)<\/strong> \u3068\u540c\u69d8\u306b\u8003\u3048\u308c\u3070\r\n\\[\r\n\\textstyle\\sum\\limits _ {k=1}^{12n} [2] = n (-1+1) = 0\r\n\\]\r\n\u3088\u3063\u3066, \u6c42\u3081\u308b\u5024\u306f\r\n\\[\\begin{align}\r\n\\textstyle\\sum\\limits _ {k=1}^{12n} {a _ k}^2 & = \\textstyle\\sum\\limits _ {1}^{12n} k^2 +2 \\cdot 6n +\\dfrac{1}{2} \\cdot 0 +\\dfrac{1}{2} \\cdot 12n \\\\\r\n& = \\dfrac{1}{6} \\cdot 12n (12n+1) (24n+1) +12n +6n \\\\\r\n& = 2n ( 288 n^2 +36n +1 +9 ) \\\\\r\n& = \\underline{4n ( 144n^2 +18n +5 )}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u6570\u5217 \\(\\{ a _ k \\}\\) \u3092 \\(a _ k = k +\\cos \\left( \\dfrac{k \\pi}{6} \\right)\\) \u3067\u5b9a\u3081\u308b. \\(n\\) \u3092\u6b63\u306e\u6574\u6570\u3068\u3059\u308b. (1)\u3000\\(\\textstyl … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n