已知数列{an}的前n项和为Sn,且Sn=2an-2(n∈N*),在数列{bn}中,b1=1,点P(bn,bn+1)在直线x-y+2=0上.(1)求数列{an}
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已知数列{an}的前n项和为Sn,且Sn=2an-2(n∈N*),在数列{bn}中,b1=1,点P(bn,bn+1)在直线x-y+2=0上. (1)求数列{an},{bn}的通项公式; (2)记Tn=a1b1+a2b2+…+anbn,求Tn. |
答案
(1)由Sn=2an-2得:Sn-1=2an-1-2(n≥2), 两式相减得:an=2an-2an-1,即=2(n≥2), 又a1=2a1-2, ∴a1=2, ∴数列{an}是以2为首项,2为公比的等比数列, ∴an=2n. ∵点P(bn,bn+1)在直线x-y+2=0上, ∴bn+1-bn=2, ∴数列{bn}是等差数列, ∵b1=1, ∴bn=2n-1; (2)Tn=1×2+3×22+5×23+…+(2n-3)×2n-1+(2n-1)×2n① ∴2Tn=1×22+3×23+…+(2n-3)×2n+(2n-1)×2n+1② ①-②得:-Tn=1×2+2(22+23+…+2n)-(2n-1)×2n+1 =2+2×-(2n-1)×2n+1 =2+2×2n+1-8-(2n-1)×2n+1 =(3-2n)2n+1-6, ∴Tn=(2n-3)2n+1+6. |
举一反三
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