高一数学题，急用啊 高一数学题急用
\u9ad8\u4e00\u6570\u5b66\u9898\uff0c\u6025\u7528\u554a\uff0c\u5feb\u5feb1\u3001\u5728\u7b49\u6bd4\u6570\u5217{an}\u4e2d\uff0ca3+a6=36\uff0ca4+a7=18\uff0can=1/2\uff0c\u6c42n\u7684\u503c\u3002
\u89e3\u6cd5\u4e00\uff1a\u8bbe\u8be5\u7b49\u6bd4\u6570\u5217\u7684\u516c\u6bd4\u4e3aq
\u5219\uff1a
a(6) = a(3) \u00d7 q³
a(7) = a(4) \u00d7 q³
\u2235 a(3) + a(6) = 36
\u2234 a(3) + a(3) \u00d7 q³ = 36
\u5373 a(3) \u00d7 ( 1 + q³ ) = 36 ----------------------------- \u2460
\u2235 a(4) + a(7) = 18
\u2234 a(4) + a(4) \u00d7 q³ = 18
\u5373 a(4) \u00d7 ( 1 + q³ ) = 18 ------------------------------ \u2461
\u2461 \u00f7 \u2460\uff0c\u5f97a(4)/a(3) = 1/2\uff0c
\u2234 \u516c\u6bd4q = 1/2
\u628aq = 1/2 \u4ee3\u5165\u2460\u5f0f\u5f97a(3) = 32
\u5219 a(1) = a(3) / q² = 128
\u628a a(1) = 128\u3001q = 1/2\u3001a(n) = 1/2
\u4ee3\u5165 a(n) = a(1) \u00d7 q\u7684(n-1)\u6b21\u65b9\uff0c\u5f97\uff1an = 9
\u89e3\u6cd5\u4e8c\uff1a\u8bbe\u7b49\u6bd4\u6570\u5217\u7684\u516c\u6bd4\u4e3aq
\u5219\uff1a
a(4) = a(3) \u00d7 q
a(7) = a(6) \u00d7 q
\u2235 a(4) + a(7) = 18
\u2234 a(3) \u00d7 q + a(6) \u00d7 q = 18
\u5373\uff1a[ a(3) + a(6) ] \u00d7 q = 18 ------------------------ \u2460
\u800c a(3) + a(6) = 36 ------------------------ \u2461
\u2460 \u00f7 \u2461 \u5f97\uff1aq = 1/2
\u2235 a(3) = a(1) \u00d7 q²
a(6) = a(1) \u00d7 q\u4e94\u6b21\u65b9
\u2234 a(3) = a(1)/4
a(6) = a(1)/32
\u800ca(3) + a(6) = 36
\u2234 a(1)/4 + a(1)/32 = 36
\u89e3\u5f97 a(1) = 128
\u628a a(1) = 128\u3001q = 1/2\u3001a(n) = 1/2
\u4ee3\u5165 a(n) = a(1) \u00d7 q\u7684(n-1)\u6b21\u65b9\uff0c\u5f97\uff1an = 9
2\u3001
\uff081\uff09\u6570\u5217{an}\u662f\u7b49\u5dee\u6570\u5217\u3002
\u8bc1\u6cd5\u4e00\uff1a
\u2235 b(n) = 2\u7684a(n)\u6b21\u65b9
\u2234 b(n+1) = 2\u7684a(n+1)\u6b21\u65b9
\u8bbe\u7b49\u6bd4\u6570\u5217{bn}\u7684\u516c\u6bd4\u4e3aq
\u5219\uff1a
q = b(n+1) / b(n)
= [ 2\u7684a(n+1)\u6b21\u65b9] / [ 2\u7684a(n)\u6b21\u65b9]
= 2\u7684 [a(n+1)\u6b21\u65b9 - a(n)\u6b21\u65b9]
\u2234 a(n+1) - a(n) = log(3)(q) = \u5b9a\u503c
\u2234 \u6240\u4ee5{an}\u662f\u7b49\u5dee\u6570\u5217
\u8bc1\u6cd5\u4e8c\uff1a
\u8bbe\u7b49\u6bd4\u6570\u5217{bn}\u7684\u516c\u6bd4\u4e3aq
\u5219\uff1a
b(n) = b(1) \u00d7 q\u7684(n-1)\u6b21\u65b9
\u800c\u5df2\u77e5 b(n) = 2\u7684(an)\u6b21\u65b9
\u2234b(1) \u00d7 q\u7684(n-1)\u6b21\u65b9 = 2\u7684(an)\u6b21\u65b9
\u2234 a(n) = log(2) [b(1) \u00d7 q\u7684(n-1)\u6b21\u65b9]
\u2234 a(n) = log(2)(b1) + log(2)[q\u7684(n-1)\u6b21\u65b9]
\u2234 a(n) = log(2)(b1) + (n-1)log(2)(q)
\u2234 \u6570\u5217{an}\u662f\u4ee5log2(b1)\u4e3a\u9996\u9879\u3001\u4ee5log(2)(q)\u4e3a\u516c\u5dee\u7684\u7b49\u5dee\u6570\u5217\u3002
\uff082\uff09
\u89e3\u6cd5\u4e00\uff1a
\u2235 {an}\u662f\u7b49\u5dee\u6570\u5217
\u2234a(1) + a(2) + a(3) + \u00b7 \u00b7 \u00b7 + a(18) + a(19) + a(20)
= 10 \u00d7 [ a(8) + a(13) ]
= 10 \u00d7 (1/2)
= 5
\u2234b1b2\u00d7\u2026\u2026\u00d7b20
= 2^[ a(1) + a(2) + a(3) + \u00b7 \u00b7 \u00b7 + a(18) + a(19) + a(20) ]
= 2\u76845\u6b21\u65b9
= 32
\u89e3\u6cd5\u4e8c\uff1a
\u2235\u7b49\u5dee\u6570\u5217{an}\u4e2d\uff0ca(8) + a(13) = 1/2 \u4e14 bn=2\u7684an\u6b21\u65b9
\u2234 b(8) \u00d7 b(13)
= [ 2\u7684(a8)\u6b21\u65b9] \u00d7 [2\u7684(a13)\u6b21\u65b9]
= 2^[a(8) + a(13)]
= 2\u7684(1/2)\u6b21\u65b9
\u5728\u7b49\u6bd4\u6570\u5217{bn}\u4e2d\uff0c
b(1) \u00d7 b(20) = b(2) \u00d7 b(19) = b(8) \u00d7 b(13) = 2\u7684(1/2)\u6b21\u65b9
\u2234b1b2\u00d7\u2026\u2026\u00d7b20
= [b(1) \u00d7 b(20)] \u00d7 [b(2) \u00d7 b(19)] \u00d7 \u00b7 \u00b7 \u00b7 \u00d7 [b(8) \u00d7 b(13)]
= 10\u4e2a[2\u7684(1/2)\u6b21\u65b9] \u7684\u4e58\u79ef
= 2\u7684[10\u00d7(1/2)] \u6b21\u65b9
= 2\u76845\u6b21\u65b9
= 32
\u8fd9\u4e9b\u90fd\u662f\u521d\u4e2d\u7684\u9898\u76ee\uff0c\u4f8b\u5982\u56e0\u5f0f\u5206\u89e3\uff0c\u548c\u7ef4\u8fbe\u5b9a\u7406\u7684\u5e94\u7528\u800c\u5df2\uff0c\u4f60\u81ea\u5df1\u5f97\u60f3\u529e\u6cd5\u505a\uff0c\u7279\u522b\u662f\u56e0\u5f0f\u5206\u89e3\uff0c\u9ad8\u4e2d\u7684\u57fa\u672c\u529f\u3002
1、解：设成等差数列的四个数分别为
a1、a2、a3、a4，公差为d。
∵四个数之和为26
∴4(a1 + a4)/2 = 26
∴a1 + a4 = 13
即 a1 + (a1 + 3d) = 13
亦即 (a1 + d) + (a1 + 2d) = 13
∴ a2 + a3 = 13 ---------------------- ①
注：在等差数列中，可直接利用 a2 + a3 = a1 + a4 = 13
∵第二个数与第三个数之积为40
∴a2 × a3 = 40 ---------------------- ②
由①②知：a2 和 a3 是方程 X² - 13X + 40 = 0 的两个实根
∴a2 = 5，a3 = 8 或 a2 = 8，a3 = 5
当 a2 = 5，a3 = 8 时，d = a3 - a2 = 3
此时a1 = a2 - d = 5 - 3 = 2，a4 = a3 + d = 8 + 3 = 11
∴ 这个等差数列为：2，5，8，11。
当 a2 = 8，a3 = 5 时，d = a3 - a2 = - 3
此时a1 = a2 - d = 8 - (- 3) = 11，a4 = a3 + d = 5 + (- 3) = 2
∴ 这个等差数列为：11，8，5，2。
综上，该等差数列为2，5，8，11 或 11，8，5，2。
2、解：
∵ a(n) = 2 - 1 / [a(n-1)]
∴ a(n) - 1 = 1 - 1 / [a(n-1)]
∴ a(n) - 1 = [a(n-1) - 1] / [a(n-1)]
上式两边取倒数
∴ 1 / [a(n) -1] = [a(n-1)] / [a(n-1) -1]
∴ 1 / [a(n) -1] = { [a(n-1) - 1] + 1} / [a(n-1) -1]
∴ 1 / [a(n) -1] = 1 + 1 / [a(n-1) -1]
即 b(n) = 1 + b(n-1)
∴ b(n) - b(n-1) = 1
∴ 数列{bn}是以1为公差的等差数列。
∵a(n) = 2 - {1 / [a(n-1)] }
∵a(n+1) = 2 - [1 / a(n)]
∴a(n+1) - 1 = 1 - [1 / a(n)] = [a(n) -1] / [a(n)]
上式两边取倒数
∴ 1 / [a(n+1) -1] = [a(n)] / [a(n) -1] = {[a(n) - 1] + 1} / [a(n) -1] = 1 + {1/[a(n) -1]}
∴ 1 / [a(n+1) -1] － {1/[a(n) -1]} ＝ 1
∴ 数列{1/[a(n) -1]}是首项为(-5/2)、公差为1的等差数列。
∴1/[a(n) -1] = (-5/2) + (n - 1) × 1 = n - (7/2) = (2n - 7)/2
∴a(n) - 1 = 2/(2n - 7)
∴a(n) = (2n - 5) / (2n - 7)
3、解：
∵△ABC中，∠A、∠B、∠C成等差数列
∴A + C = 2B
而A + C = 180° - B
∴B = 60°
∴A + C = 120°
∴A = 120° - C
∴A - C = 120° - 2C
而 0° ≤ (120° - 2C) < 120°
∴ 0 ≤ sin(120° - 2C) ≤ 1
cos²A + cos²C
= (1 + cos2A)/2 + (1 + cos2C)/2
= (2 + cos2A + cos2C)/2
= 1 + (cos2A + cos2C)/2
= 1 + [2cos(A + C) sin(A - C)]/2
= 1 + cos(A + C) sin(A - C)
= 1 - cosBsin(A - C)
= 1 - cos60° × sin(A - C)
= 1 - (1/2) × sin(A - C)
= 1 - (1/2) × sin(120° - 2C)
而 0 ≤ sin(120° - 2C) ≤ 1 （已证）
∴ 1 - (1/2) × sin(120° - 2C) ∈ [1/2，1]
∴ cos²A+cos²C取值范围是 [1/2，1]。
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绛旓細1.褰搙=-1鏃讹紝B={1,x+1,x^2-2x+2,-1/2(x^2-3x-8),x^3+x^2+3x+7}={1,0,5,2,4} A鈭〣={2,4,5}锛屾墍浠=-1鑸嶅幓銆2.褰搙=1鏃讹紝B={1,x+1,x^2-2x+2,-1/2(x^2-3x-8),x^3+x^2+3x+7}={1,2,5,12} 婊¤冻A鈭〣={2,5}銆3.褰搙=2鏃讹紝B={1,x+1,x...
绛旓細a3=a1q^2=2 1 S4=5S2 a1(q^4-1)/(q-1)=5a1(q^2-1)/(q-1)(q^2+1)(q^2-1)=5(q^2-1)q^2+1=5 q^2=4 鍥犱负鍏瘮q灏忎簬1 鎵浠=-2 浠ｅ叆1寮忓緱 a1=1/2 An=a1q^(n-1)=1/2*(-2)^(n-1)=2^(-1)*(-2)^(n-1)=-(-2)^(-1)*(-2)^(n-1)=-(-2...
