急 F6 Arithmetic Sequences 1q15

34a)F(3)= (3)^4 -8(3)^3 +14(3)^2 +8(3)-15
= 81-216+126+24-15
=0
b)因為4個solutions of the equation for an arithmetic sequence with common difference 2
所以 ans會向 f(3 ±2) 方向走
可試 先 f(3-2)走先
f(1)=1-8+14+8-15
=0
所以f(1)是其中一個solution
在試f(-1) =1-8(-1)+14+8(-1)-15
=1+8+14-8-15
=0
所以f(-1)都是solution,繼續
f(-3)=81-8(-27)+14(9)+8(-3)-15
=81+216+126-24-15
=384
因為f(-3)不等於0 ,所以f(-3)不是solution
而剩下一個就是f(3+2)=f(5)
所以f(x)
x=5,3,1,-1

35a)
T1 =log a [r^(1-1)]
T2 =log a [r^(2-1)]
T3 =log a [r^(3-1)]
.....
so
Tn = log a [r^(n-1)]

b)log ar-log a =(log a +log r) -log a =log r
log ar^2 -log ar = (log a + 2log r) - (log a +log r) =log r
........
so : log a , log ar , log ar^2 , .... is an arithmetic sequence.

c) from (b) ... the common difference of the sequence is log r.

d) the fisrt 20 terms of the sequence is :

n
-- [ 2a + (n -1 ) d ]
2

= 20/2 [2(log0.01) + (20-1) log 10]
= 10 [2 (-2) +19 (1)]
= 10 [ -4+19]
=10 (15)
=150