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=anb1+an-1b2+...+a1bn,证明Tn+12=-2an+10bn (n∈N+)最好可以是写在纸上拍下来的答案,方
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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=anb1+an-1b2+...+a1bn,证明Tn+12=-2an+10bn (n∈N+)最好可以是写在纸上拍下来的答案,方](/uploads/image/z/1549922-50-2.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%281%29+%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%882%EF%BC%89%E8%AE%B0Tn%3Danb1%2Ban-1b2%2B...%2Ba1bn%2C%E8%AF%81%E6%98%8ETn%2B12%3D-2an%2B10bn+%EF%BC%88n%E2%88%88N%2B%EF%BC%89%E6%9C%80%E5%A5%BD%E5%8F%AF%E4%BB%A5%E6%98%AF%E5%86%99%E5%9C%A8%E7%BA%B8%E4%B8%8A%E6%8B%8D%E4%B8%8B%E6%9D%A5%E7%9A%84%E7%AD%94%E6%A1%88%2C%E6%96%B9)
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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=anb1+an-1b2+...+a1bn,证明Tn+12=-2an+10bn (n∈N+)最好可以是写在纸上拍下来的答案,方
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=anb1+an-1b2+...+a1bn,证明Tn+12=-2an+10bn (n∈N+)
最好可以是写在纸上拍下来的答案,方便思维嘛.
只求第二问
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=anb1+an-1b2+...+a1bn,证明Tn+12=-2an+10bn (n∈N+)最好可以是写在纸上拍下来的答案,方
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
=
12(1-2 n-1)
1-2
+2n+2-6n+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*).
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