已知{an}是公差为d的等差数列,其前n项和为Sn,{bn}是公比为q的等比数列,且a1=b1=3,a3=b2-2,S4=b3-3.
2个回答
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(1)由a1=b1=3,知a3=3+2d,b2=3q,
S4=
(3+3+3d)×4
2
=12+6d,b3=3q2,
从而可得
3+2d=3q?2
12+6d=3q2?3
,即
2d=3q?5
6d=3q2?15
,
解得q=3,d=2,
从而有an=3+(n-1)×2=2n+1,bn=3×3n?1=3n.
(2)由(1)可知cn=anbn=(2n+1)×3n,
∴Tn=c1+c2+…+cn=3×3+5×32+7×33+…+(2n+1)×3n,
则3Tn=3×32+5×33+7×34+…+(2n-1)×3n+(2n+1)×3n+1,
两式相减得-2Tn=3×3+2×32+2×33+…+2×3n-(2n+1)×3n+1
=3×3+2×32(1+3+32+33+…+3n-2)-(2n+1)×3n+1
=9+18×
3n?1?1
2
-(2n+1)×3n+1,
∴Tn=n×3n+1.
S4=
(3+3+3d)×4
2
=12+6d,b3=3q2,
从而可得
3+2d=3q?2
12+6d=3q2?3
,即
2d=3q?5
6d=3q2?15
,
解得q=3,d=2,
从而有an=3+(n-1)×2=2n+1,bn=3×3n?1=3n.
(2)由(1)可知cn=anbn=(2n+1)×3n,
∴Tn=c1+c2+…+cn=3×3+5×32+7×33+…+(2n+1)×3n,
则3Tn=3×32+5×33+7×34+…+(2n-1)×3n+(2n+1)×3n+1,
两式相减得-2Tn=3×3+2×32+2×33+…+2×3n-(2n+1)×3n+1
=3×3+2×32(1+3+32+33+…+3n-2)-(2n+1)×3n+1
=9+18×
3n?1?1
2
-(2n+1)×3n+1,
∴Tn=n×3n+1.
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