设数列{an}的前n项和为Sn=2n^2,{bn}为等比数列,且a1=b1,b2(a2-a1)=b1.

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  • a1=S1=2

    Sn=2n^2

    Sn-1=2(n-1)^2=2n^2-4n+2

    an=Sn-Sn-1=2n^2-2n^2+4n-2=4n-2

    n=1代入4-2=2=a1,同样满足.

    数列{an}通项公式为an=4n-2

    b1=a1=2

    a2=4×2-2=6

    b2(a2-a1)=b1

    b2(6-2)=2

    b2=1/2

    b2/b1=(1/2)/2=1/4

    数列{bn}是以2为首项,1/4为公比的等比数列.

    bn=2(1/4)^(n-1)=8/4^n

    数列{bn}的通项公式为bn=8/4^n

    cn=an/bn=(4n-2)/[8/4^n]=(2n-1)4^n/4=2n4^(n-1)-4^(n-1)

    Tn=2[1×4^0+2×4^1+3×4^2+...+n×4^(n-1)]-(4^n-1)/(4-1)

    令Mn=1×4^0+2×4^1+3×4^2+...+n×4^(n-1)

    则4Mn=4^1+2×4^2+3×4^3+...+(n-1)×4^(n-1)+n×4^n

    Mn-4Mn=-3Mn=4^0+4^1+4^2+...+4^(n-1)-n×4^n=(4^n-1)/(4-1)-n4^n

    Mn=n4^n/3-(4^n-1)/9

    Tn=2n4^n/3-2(4^n-1)/9-(4^n-1)/3

    =[6n4^n-2×4^n+2-3×4^n+3]/9

    =[(6n-2-3)4^n+5]/9

    =[(6n-5)4^n+5]/9