已知无穷等比数列{an}的{bn}的公比的绝对值都小于1,且a1=b1......

1. C
{an} 的前n项和Sn = a1(1-q^n)/(1-q)，【q^n意为q的n次方】
当n趋近于无穷大时，如果q>0，Sn的极限为a1/(1-q)，如果q<0，则该极限不存在。由题可知当n趋近于无穷大时Sn的极限为3a1，所以q = 2/3。设{bn}的公比为p，由b3=a5，a1=b1有
p^2 = q^4 ，解得 p = ±4/9，又因为{bn}中至少有一项与{an}的各项都不相同，所以
p= -4/9
2. Sn = (1/d) × (1/a1 - 1/a2 + 1/a2 - 1/a3 + 1/a3 - 1/a4 + ……+ 1/an - 1/an+1)
= (1/d) × (1/a1 - 1/an+1)
= (1/d) × (1/a1 - 1/(n×d×a1))
当n趋近无穷大时 lim1/(n×d×a1) = 0
所以limSn = 1/（d×a1 ）
3. a=1，b=0
lim[(n^3+2n+1)/(n^2+3) -an-b] = -b + lim[(n^3+2n+1)/(n^2+3) - an(n^2 + 3))/(n^2+3)]
= -b + lim[(1-a)(n^3)+(2-3a)n+1]/(n^2 + 3) = 0
只有当（1-a）= 0 时极限存在，否则极限为无穷大，所以a=1.
将a=1带入上式得：-b + lim(1-n)/(n^2 + 3) = 0
-b + 0 = 0
解得a=1，b=0
4. 1
5, C
原式= lim n*(2/3)*(3/4)*(4/5)*……*((n+1)/(n+2))
= lim 2n/(n+2)
= lim 2/(1+2/n)
= 2
6. 因为1/n(n+1)(n+2)=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}
则原式=lim 1/2[1/1*2-1/2*3+1/2*3-1/3*4……+1/n(n+1)-1/(n+1)(n+2)]
=lim 1/2[1/1*2-1/(n+1)(n+2)]
= 1/4
7. 原式的分子分母同除以b^n:
= lim [(a/b)^n + p + (1/b)^n]/[a*(a/b)^n - 2 + 3/b^n] 【 因为1 所以lim(a/b)^n=0，lim3/b^n=0】
= - 2/p = 2
p = -1

a3=a1+2d a5=a1+4d
b3=b1*d^2 b5=b1*d^4
a1+2d=b1*d^2 a1+4d=5b1*d^4
2d=3a1*d^2-a1 (1) 4d=a1*d^4-a1 (2)
(2)/(1)
2=(5d^4-1)/(3d^2-1)
5d^4-6d^2+1=0
d^2=1 (舍去) d^2=1/5
d=√5/5 a1=-√5
an=-√5+(n-1)*√5/5
bn=-√5*(√5/5)^(n-1)

a3=a1+2d a5=a1+4d
b3=b1*d^2 b5=b1*d^4
a1+2d=b1*d^2 a1+4d=5b1*d^4
2d=3a1*d^2-a1 (1) 4d=a1*d^4-a1 (2)
(2)/(1)
2=(5d^4-1)/(3d^2-1)
5d^4-6d^2+1=0
d^2=1 (舍去) d^2=1/5
d=√5/5 a1=-√5
an=-√5+(n-1)*√5/5
不知道