已知{an}是等差数列,若其前n项和为Sn,{bn}等比数列,且a1=b1,a4+b4=27,S4-b4=10,求数列{

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  • ∵a4+b4=27,s4-b4=10 ∴a4+S4=37 ∴a4+2a1+2a4=37 ∴2a1+3a4=37

    ∴5a1+9d=37 ∴9d=27 ∴d=3 ∴an=a1+(n-1)d=3n-1

    ∵a4+b4=27 ∴11+2q³=27 ∴q³=8 ∴q=2 ∴bn=b1q^(n-1)=2^n

    ∵Tn=anb1+an-1b2+...+a1bn ∴2Tn=anb2+an-1b3+...+a2bn+a1bn+1

    两式相减得:Tn=(an-an-1)b2+(an-1-an-2)b3+...+(a2-a1)bn+a1bn+1-anb1

    =3(b2+b3+...+bn)+a1bn+1-anb1

    =3×2²[2^(n-1)-1]+2×2^(n+1)-2an

    =3×2×2^n-12+4×2^n-2an

    =6bn-12+4bn-2an

    ∴Tn=10bn-12-2an 即 Tn+12=﹣2an+10bn