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+b4=27,S4-b4=10（1）求数列{an}与{bn}的通项公式；（2）记Tn=a1b1+a2b2+...+anbn,n∈N+,证明Tn-8=an-1bn+1

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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+b4=27,S4-b4=10（1）求数列{an}与{bn}的通项公式；（2）记Tn=a1b1+a2b2+...+a

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+b4=27,S4-b4=10（1）求数列{an}与{bn}的通项公式；（2）记Tn=a1b1+a2b2+...+anbn,n∈N+,证明Tn-8=an-1bn+1
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+b4=27,S4-b4=10
（1）求数列{an}与{bn}的通项公式；（2）记Tn=a1b1+a2b2+...+anbn,n∈N+,证明Tn-8=an-1bn+1

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+b4=27,S4-b4=10（1）求数列{an}与{bn}的通项公式；（2）记Tn=a1b1+a2b2+...+anbn,n∈N+,证明Tn-8=an-1bn+1
(1)设数列{an}的公差是d,{bn}的公比是q,依题意
2+3d+2q^3=27,①
8+6d-2q^3=10,②
①+②,10+9d=37,d=3,
代入①,11+2q^3=27,q^3=8,q=2.
∴an=2+3(n-1)=3n-1,
bn=2^n.
(2)Tn=2*2+5*2^2+8*2^3+……+（3n-1)*2^n,③
∴2Tn= 2*2^2+5*2^3+……+（3n-4)*2^n+(3n-1)*2^(n+1),④
③-④,-Tn=4+3(2^2+2^3+……+2^n)-(3n-1)*2^(n+1)
=4-3[2^2-2^(n+1)]-(3n-1)*2^(n+1),
=-8-(3n-4)*2^(n+1),
∴Tn=8+(3n-4)*2^(n+1),
∴Tn-8=ab.

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