已知等差数列{an}的首项a1=1,公差d>0,数列{bn}是等比数列,且满足a1=b1,a2=b2,a5=b3

an = 1+(n-1)d
bn = b1q^(n-1)
a1=b1 =1
a2=b2
1+d =q (1)
a5=b3
1+4d = q^2 (2)
(1)^2 /(2)
(1+d)^2 = 1+4d
d^2-2d =0
d= 2
q= 3
ie
an = 1+(n-1)2 = 2n-1
bn=b1q^(n-1) = 3^(n-1)
c1/b1+c2/b2+……+cn/bn=a(n+1) (3)
c1/b1+c2/b2+……+c(n-1)/b(n-1)=an (4)
(3)-(4)
cn/bn = a(n+1) -an
cn = bn.[a(n+1) -an]
= 2.3^(n-1)
c1+c2+...+c2013
=(3^2013)- 1

（1）∵a2=1+d，a5=1+4d，a14=1+13d，且a2、a5、a14成等比数列
∴（1+4d）2=（1+d）（1+13d）即d=2∴an=1+（n-1）•2=2n-1
又∵b2=a2=3，b3=a5=9、∴q=3，b1=1，bn=3n-1
（2）∵
C1
b1
+
C2
b2