6.(2012•天津)已知{an}是等差数列,其前n项和为Sn,{bn}是等比数列,且a1=b1=2,a4+

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  • (1)设等差数列的公差为d,等比数列的公比为q,

    由a1=b1=2,得a4=2+3d,b4=2q3,S4=8+6d,

    由条件a4+b4=27,S4-b4=10,

    +2n+2-6n+2

    故an=3n-1,bn=2n,n∈N*.

    (2)由(1)得,Tn=2an+22an-1+23an-2+…+2na1;①;

    2Tn=22an+23an-1+…+2na2+2n+1a1;②;

    由②-①得,Tn=-2(3n-1)+3×22+3×23+…+3×2n+2n+2 =10×2n-6n-10;

    而-2an+10bn-12=-2(3n-1)+10×2n-12=10×2n-6n-10;

    故Tn+12=-2an+10bn(n∈N*).