Question

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

Answer 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=a<n-1>b<n+1>.

Answer 2

∵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