已知数列{an}满足an=2an-1+2n-1(n≥2),a1=5,bn=an−12n
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已知数列{an}满足an=2an-1+2n-1(n≥2),a1=5,bn=
a
(I)证明:∵an=2an-1+2n-1(n≥2),∴an−1=2(an−1−1)+2n,
∴ an−1 2n= an−1−1 2n−1+1.∴bn=bn-1+1. ∴{bn}是首项为 a1−1 2= 5−1 2=2,公差为1的等差数列; (II)由(I)可得bn=2+(n-1)×1=n+1, ∴ an−1 2n=n+1,∴an=(n+1)•2n+1, 令cn=(n+1)•2n,其前n项和为Tn, 则Tn=2×2+3×22+4×23+…+n•2n-1+(n+1)•2n, 2Tn=2×22+3×23+…+n•2n+(n+1)•2n+1, 两式相减得-Tn=2×2+22+23+…+2n-(n+1)•2n+1=2+ 2(2n−1) 2−1-(n+1)•2n+1=-n•2n+1, ∴Tn=n•2n+1. ∴Sn=Tn+n=n+n•2n+1.
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