1.等差数列的第一项是2,第二项是5,那么第八项是?2.a为125和5的等比中项,则a=?3.若x,y,z成等差数列,则(z-x)平方+4(x-y)(y-x)=?4.在等比数列中,S10=10,S20=15,则S30=?5.数列1/1*3,1/3*5,1/5*7,…,1/(2n-1)(2n+1)的前n项

Question

1.等差数列的第一项是2,第二项是5,那么第八项是?
2.a为125和5的等比中项,则a=?
3.若x,y,z成等差数列,则(z-x)平方+4(x-y)(y-x)=?
4.在等比数列中,S10=10,S20=15,则S30=?
5.数列1/1*3,1/3*5,1/5*7,…,1/(2n-1)(2n+1)的前n项和为多少?

Answer 1

1.由题设公差 d=a2-a1=3
∴ a8=a1+7d=2+7·3=23
2.由 125:a=a:5,解得 a=±√5×125=±25
3.设公差为d
则 y-x = z-y = d,
z-x = 2d
原式 = 4d²+4(-d)·d = 0
4.由等比级数前n项和的公式Sn=a1(1-q^n)/(1-q)得
10=S10=a1(1-q^10)/(1-q) ---(1)
15=S20=a1(1-q^20)/(1-q) ---(2)
(2)/(1)得3/2=S20/S10=（1-q^20)/(1-q^10) ---(3)
整理得 2(q^10)²-3(q^10)+1=0
∴ q^10=1(不合题意舍去),
q^10=1/2
类似(3)式：
S30/S10=(1-q^30)/(1-q^10)
=[1-(q^10)³]/(1-q^10)
=[1-(1/2)³]/[1-1/2]
=7/4
∴ S30=7/4·S10=35/2
5.1/(2n-1)(2n+1)=[1/(2n-1)-1/(2n+1)]/2
1/1*3+1/3*5+1/5*7+…+1/(2n-1)(2n+1)
=(1/2)[1/1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)