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.
∵ 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)
∵△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].
