數學知識交流 － 數學雜題 (1)

(1) 求數列1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, ... 的
(a) 通項。 (b) 第四、七、十三及二十七項。
解：這是有名的卡塔蘭數，通項是Cn = (2n)! / [(n+1)! n!]。
第四項是：C4 = (2×4)! / [(4+1)! × 4!] = 8! / (5! × 4!) = 14.
第七項是：C7 = (2×7)! / [(7+1)! × 7!] = 14! / (8! × 7!) = 429.
第十三項是：C13 = (2×13)! / [(13+1)! × 13!] = 26! / (14! × 13!) = 742900.
第二十七項是：C27 = (2×27)! / [(27+1)! × 27!] = 54! / (28! × 27!) = 69533550916004.(2a) 解方程 56x³ + 218x² + 17x - 210 = 0。
解：56x³ + 218x² + 17x - 210 = 0
8x²-----38x------35
-----7x---------6
(8×7 = 56，8×(-6) + 38×7 = 218，35×7 + 38×(-6) = 17，35×(-6) = -210)
(8x² + 38x + 35)(7x - 6) = 0
(2x + 7)(4x + 5)(7x - 6) = 0
∴ x1 = -7/2，x2 = -5/4，x3 = 6/7.

(2b) 從 (2a)，解方程 56(y³-3y²+3y-1) + 218(y²-2y+1) + 17(y-1) - 210。
解：56(y³-3y²+3y-1) + 218(y²-2y+1) + 17(y-1) - 210 = 0
56(y-1)³ + 218(y-1)² + 17(y-1) - 210 = 0
把(y-1)看作(2a)中的x，則
y - 1 = -7/2 或 -5/4 或 6/7
∴ y1 = -5/2，y2 = -1/4，y3 = 13/7.(3) 求 g[f(m)] 若 f(x) = (x+m)(x+6) , f(9) = 30 , g(x) = 8(8+x)。
解：f(x) = (x+m)(x+6)
f(9) = (9+m)(9+6) = 15(9+m) = 30，m = -7
so f(x) = (x-7)(x+6) and g(x) = 8(x+8)
so g[f(m)]
= g[(m-7)(m+6)]
= g[(-7-7)(-7+6)]
= g(14)
= 8(14+8)
= 176(4) 求
(a) 7! = 7×6×5×4×3×2×1 = 5040
(b) 8! - 7! = 7! × 8 - 7! = 5040 × 8 - 5040 = 35280
(c) (8!-7!)/4! = 35280 / (4×3×2×1) = 1470
(d) 2!! = 2
(e) 3!! = 3 × 1 = 3
(f) 5!! = 5 × 3 × 1 = 15
(g) 8!! = 8 × 6 × 4 × 2 = 384
(h) 0!! = 1
(i) (-1)!! = 1