第 13 題
寫出幾項 a_n，可猜到 2^(n + 2) - 7(a_n)^2 = [2a_(n + 1) + a_n]^2
然後用數學歸納法證明

(1) n = 1 時，2^3 - 7(a_1)^2 = (2a_2 + a_1)^2 = 1

(2) 設 n = k 時，2^(k + 2) - 7(a_k)^2 = [2a_(k + 1) + a_k]^2

(3) n = k + 1 時
2^(k + 3) - 7[a_(k + 1)]^2
= 2{7(a_k)^2 + [2a_(k + 1) + a_k]^2} - 7[a_(k + 1)]^2
= [a_(k + 1)]^2 + 8[a_(k + 1)](a_k) + 16(a_k)^2
= [a_(k + 1) + 4(a_k)]^2
= [-2a_(k + 1) - 4(a_k) + a_(k + 1)]^2
= [2a_(k + 2) + a_(k + 1)]^2

第 5 題
(1) 任意分，H(3，3) * H(3，4) * H(3，5)
(2) 恰有一人沒分到，C(3，1) * H(2，3) * H(2，4) * H(2，5)
(3) 恰有兩人沒分到，3 種
所求 = (1) - (2) + (3)

第 11 題
x < 0，f(x / (x - 1)) = xf(x)
x = -1，f(1/2) = -f(-1) = f(1) = 1
x = -1/2，f(1/3) = (-1/2)f(-1/2) = (1/2)f(1/2) = 1/2!
x = -1/3，f(1/4) = … = 1/3!
:
:
f(1/k) = 1/(k - 1)!

所求 = 1/99! + 1/98! + 1/2! * 1/97! + … + 1/49! * 1/50!
= (1/99!) * {[C(99，0) + C(99，99)] / 2 + [C(99，1) + C(99，98)] / 2 + … + [C(99，49) + C(99，50)] / 2}
= 2^98 / 99!

第 12 題
令 g(x) = sin(πx) - cos(πx) + 2 = √2sin(πx - π/4) + 2，其中 1/4 <= x <= 5/4
若 1/4 <= a <= 3/4，必存在 3/4 <= b <= 5/4，使得 g(a) = g(b)
f(a) = g(a) / √a = g(b) / √a > g(b) / √b = f(b)
又 f(x) = g(x) / √x 在 3/4 <= x <= 5/4 遞減
所求 = f(5/4) = (4/5)√5

[ 本帖最後由 thepiano 於 2025-9-16 15:43 編輯 ]