\u5b9f\u6570 \\(x\\) \u306b\u5bfe\u3057,\r\n\\(x\\) \u3092\u8d85\u3048\u306a\u3044\u6700\u5927\u306e\u6574\u6570\u3092 \\([ x ]\\) \u3067\u8868\u3059.\r\n\u6570\u5217 \\(\\{ a_k \\}\\) \u3092\r\n\\[\r\na_k = 2^{\\left[ \\sqrt{k} \\right]} \\quad ( k = 1, 2, 3, \\cdots )\r\n\\]\r\n\u3067\u5b9a\u7fa9\u3059\u308b.\r\n\u6b63\u306e\u6574\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066\r\n\\[\r\nb_n = \\textstyle\\sum\\limits _ {k=1}^{n^2} a_k\r\n\\]\r\n\u3092\u6c42\u3081\u3088.<\/p>\r\n
\u6574\u6570 \\(m\\) \u306b\u3064\u3044\u3066, \\(m^2 \\leqq k \\leqq (m+1)^2 -1 = m^2 +2m \\cdots \\maru{1}\\) \u306e\u3068\u304d, \\(\\left[ \\sqrt{k} \\right] = m\\) .
\r\n[1] \u3092\u307f\u305f\u3059\u6574\u6570 \\(k\\) \u306f\r\n\\[\r\n( m^2 +2m ) -m^2 +1 = 2m+1 \\ \\text{\u500b}\r\n\\]\r\n\u3042\u308b.
\r\n\u3057\u305f\u304c\u3063\u3066, \\(b_1 = 2\\) \u3067\u3042\u308a, \\(n \\geqq 2\\) \u306b\u3064\u3044\u3066\r\n\\[\\begin{align}\r\nb_n & = \\underline{\\textstyle\\sum\\limits _ {m=1}^{n-1} (2m+1) 2^m} _ {[2]} +2^n\r\n\\end{align}\\]\r\n\\(S_n = [2]\\) \u3068\u304a\u304f\u3068\r\n\\[\\begin{array}{rllrllll}\r\nS_n & = & 3 \\cdot 2 & +5 \\cdot 2^2 & +\\cdots & +(2n-1) 2^{n-1} & & ... [3] \\\\\r\n2 S_2 & = & & 3 \\cdot 2^2 & +\\cdots & +(2n-3) 2^{n-1} & +(2n-1) 2^n & ... [4]\r\n\\end{array}\\]\r\n\\([4] -[3]\\) \u3088\u308a\r\n\\[\\begin{align}\r\nS_n & = (2n-1) 2^n -2 \\left( 2^2 +\\cdots +2^{n-1} \\right) -3 \\cdot 2 \\\\\r\n& = (2n-1) 2^n -8 \\cdot \\dfrac{2^{n-2} -1}{2-1} -6 \\\\\r\n& = (2n-1) 2^n -2^{n+1} +2 \\\\\r\n& = (2n-3) 2^n +2\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\r\nb_n = \\underline{(n-1) 2^{n+1} +2}\r\n\\]\r\n\u3053\u308c\u306f, \\(n = 1\\) \u306e\u3068\u304d\u3082\u6210\u7acb\u3059\u308b.<\/p>\r\n","protected":false},"excerpt":{"rendered":"\u5b9f\u6570 \\(x\\) \u306b\u5bfe\u3057, \\(x\\) \u3092\u8d85\u3048\u306a\u3044\u6700\u5927\u306e\u6574\u6570\u3092 \\([ x ]\\) \u3067\u8868\u3059. \u6570\u5217 \\(\\{ a_k \\}\\) \u3092 \\[ a_k = 2^{\\left[ \\sqrt{k} \\right]} \\quad ( … \u7d9a\u304d\u3092\u8aad\u3080