已知数列{an}满足2an+1=an+an+2(n∈N+),其前n项和为Sn,{bn}是等比数列,且a1=b1=2,a4+b4=27,S4-b4=
已知数列{an}满足2an+1=an+an+2(n∈N+),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...
已知数列{an}满足2an+1=an+an+2(n∈N+),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+,求Tn.
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(1)由2an+1=an+an+2(n∈N+),知{an}为等差数列,设等差数列的公差为d,等比数列{bn}的公比为q,
由a1=b1=2,得a4=2+3d,b4=2q3,s4=8+6d,
由a4+b4=27,S4-b4=10,得方程组
,
解得
,
所以:an=3n-1,bn=2n.
(2)由(Ⅰ)知an?bn=(3n-1)?2n,
Tn=a1b1+a2b2+…+anbn,
则Tn=2×2+5×22+8×23+…+(3n-1)×2n,①;
2Tn=2×22+5×23+…+(3n-4)×2n+(3n-1)×2n+1,②.
由①-②得,-Tn=2×2+3×22+3×23+…+3×2n-(3n-1)×2n+1
=
-(3n-1)×2n+1-2
=-(3n-4)×2n+1-8.
所以Tn=(3n-4)×2n+1+8.
由a1=b1=2,得a4=2+3d,b4=2q3,s4=8+6d,
由a4+b4=27,S4-b4=10,得方程组
|
解得
|
所以:an=3n-1,bn=2n.
(2)由(Ⅰ)知an?bn=(3n-1)?2n,
Tn=a1b1+a2b2+…+anbn,
则Tn=2×2+5×22+8×23+…+(3n-1)×2n,①;
2Tn=2×22+5×23+…+(3n-4)×2n+(3n-1)×2n+1,②.
由①-②得,-Tn=2×2+3×22+3×23+…+3×2n-(3n-1)×2n+1
=
6×(1?2n) |
1?2 |
=-(3n-4)×2n+1-8.
所以Tn=(3n-4)×2n+1+8.
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