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设数列{an}满足a1=2,an+1=an+3•2n-1.

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设数列{an}满足a1=2,an+1=an+3•2n-1.
(1)∵a1=2,an+1-an=3•2n-1,
∴an=a1+(a2-a1)+(a3-a2)+…+(an-an-1
=2+3×20+3×21+3×22+…+3×2n-2
=2+3(20+21+22+…+2n-2
=2+3×
1(1-2n-1)
1-2=3×2n-1-1(n≥2),
经验证n=1也成立,∴an=3×2n-1-1;
(2)bn=nan=3n×2n-1-n,
b1=3×1•20-1,b2=3×2•21-2,b3=3×3•22-3,…,bn=3n•2n-1-n,
∴Sn=3[(1•20+2•21+3•22+…+n•2n-1)-(1+2+3+…+n)],
设x=1•20+2•21+3•22+…+n•2n-1①,则2x=1•21+2•22+3•23+…+n•2n②,
①-②得,-x=1+21+22+23+…+2n-1-n•2n
=1+
2(1-2n-1)
1-2-n•2n=-1+(1-n)•2n
∴x=(n-1)2n+1,
∴Sn=3[(n-1)2n+1-
n(n+1)
2],
∴Sn=(3n-3)•2n-
3
2n(n+1)+3;
(3)∵an=3×2n-1-1;
∴cn=log22n-1=n-1,

1
c2c3+
1
c3c4+…+
1
cncn+1=
1
1×2+
1
2×3+…+
1
(n-1)n
=1-
1
2+
1
2-
1
3+…+
1
n-1-
1
n=1-
1
n<1.