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+...+anbn,n∈N+,证明Tn-8=an-1bn+1](/uploads/image/z/2430642-66-2.jpg?t=%E5%B7%B2%E7%9F%A5%7Ban%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%7Bbn%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E4%B8%94a1%3Db1%3D2%2Ca4%2Bb4%3D27%2CS4-b4%3D10%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%8E%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%EF%BC%882%EF%BC%89%E8%AE%B0Tn%3Da1b1%2Ba2b2%2B...%2Banbn%2Cn%E2%88%88N%2B%2C%E8%AF%81%E6%98%8ETn-8%3Dan-1bn%2B1)
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.