Question

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

设数列{cn}对任意n均有c1/b1+c2/b2+……+cn/bn=a（n+1），求c1+c2+……+c2013的值

Answer 1

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

Answer 2

（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
++
Cn
bn
=an+1①
∴
C1
b1
=a2即C1=b1a2=3
又
C1
b1
+
C2
b2
++
Cn-1
bn-1
=an (n≥2)②
①-②：
Cn
bn
=an+1-an=2
∴Cn=2•bn=2•3n-1（n≥2）
∴Cn=
3 (n=1)
2•3n-1 (n≥2)

∴
C1+C2+C 3
++C2010=3+2•31+2•32++2•32013-1

=3+2•(31+32+33++32012)=3+2•3(1-32012)
1-3
=32013