✔ 最佳答案
(1)
Simple calculation shows that the ten terms are:
p, q, p+q, p+2q, 2p+3q, 3p+5q, 5p+8q (= 181), 8p+13q, 13p+21q, 21p+34q
The sum of these 10 terms are 55p+88q.【Check by yourself】
Note that we are lucky enough that the sum
= 55p+88q
= 11 (5p+8q)
= 11 x 7th term
= 11 x 181
= 1991
(2)
The way I do this question is try to "make" x²+5x-6 appear in the first term.
(x²+3x+2)(x²+7x+12)+(x²+5x-6)=0
(x²+5x-6-2x+8)(x²+5x-6+2x+18)+(x²+5x-6)=0
[(x²+5x-6)+(-2x+8)][(x²+5x-6)+(2x+18)]+(x²+5x-6)=0
(x²+5x-6)² + (2x+18)(x²+5x-6) + (-2x+8)(x²+5x-6) + (-2x+8)(2x+18) + (x²+5x-6) = 0
(x²+5x-6)² + (2x+18-2x+8)(x²+5x-6) + (-4x²-36x+16x+144) + (x²+5x-6) = 0
(x²+5x-6)² + 26(x²+5x-6) + (-3x²-15x+138) = 0
(x²+5x-6)² + 26(x²+5x-6) + (-3)(x²+5x-6) + 120 = 0
(x²+5x-6)² + 23(x²+5x-6) + 120 = 0
[(x²+5x-6)+15][(x²+5x-6)+8] = 0
(x²+5x+9)(x²+5x+2) = 0
x²+5x+9=0 or x²+5x+2=0
x={-5±√[5²-4(1)(9)]}/2 or x={-5±√[5²-4(1)(2)]}/2
x=(-5±√11 i)/2 or x=(-5±√17)/2 where i=√(-1) is the imaginary unit
(3)
I think there's some bad notation in you question. The recursive definition should be "(b+1)/a" but not "b+1/a"
(a)
Let an denotes the n-th term.
By listing the first few terms, we have,
a1 = 1
a2 = 3
a3 = (3+1)/1 = 4
a4 = (4+1)/3 = 5/3
a5 = (5/3 + 1) / 4 = (8/3) / 4 = 2/3
a6 = (2/3 + 1) / (5/3) = (5/3) / (5/3) = 1
a7 = (1+1) / (2/3) = 2 / (2/3) = 3
So the terms are periodic with a1 = a6 = a11 = a16 = ...
Since 100 = 19 x 5 + 5, a100 = a5 = 2/3
(b)
The sum from a1 to a5 = 1 + 3 + 4 + 5/3 + 2/3 = 31/3
In the first 152 terms, there are totally 30 sets of (a1 to a5)
So the sum of first 152 terms
= 30 x 31/3 + a151 + a152
= 310 + a1 + a2
= 310 + 1 + 3
= 314
Hope it helps! ^^
2006-12-28 11:59:46 補充：
4.Let α and 5α be the roots.Then α 5α = -b/aα = -b/(6a) ---(1)and α(5α) = c/a5α² = c/a ---(2)Sub. (1) into (2),5[-b/(6a)]² = c/a5b² = c/a * 36a²5b² = 36acThe answer is (a).
