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问:记Tn=anb1+an-1b2+...+a1bn,证明Tn+12=-2an+10bn (n∈N+)
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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问:记Tn=anb1+an-1b2+...+a1bn,证明Tn+12=-2an+10bn (n∈N+)](/uploads/image/z/2707774-70-4.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%E9%97%AE%3A%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)
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问:记Tn=anb1+an-1b2+...+a1bn,证明Tn+12=-2an+10bn (n∈N+)
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
问:记Tn=anb1+an-1b2+...+a1bn,证明Tn+12=-2an+10bn (n∈N+)
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问:记Tn=anb1+an-1b2+...+a1bn,证明Tn+12=-2an+10bn (n∈N+)
∵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
a1=b1=2为啥呀