(2014•东营二模)已知数列{an}为等差数列,且a5=14,a7=20,数列{bn}的前n项和为Sn,且满足3Sn=
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(2014•东营二模)已知数列{an}为等差数列,且a5=14,a7=20,数列{bn}的前n项和为Sn,且满足3Sn=Sn-1+2(n≥2,n∈N*),b
(1)∵数列{an}是等差数列,设公差为d,
∵a5=14,a7=20,∴
a1+4d=14
a1+6d=20,解得
a1=2
d=3,
∴an=a1+(n-1)d=3n-1.(2分)
∵3Sn=Sn-1+2(n≥2)①,
∴3Sn-1=Sn-2+2(n≥3)②,
由①-②得3bn=bn-1(n≥3),
∴
bn
bn−1=
1
3(n≥3),(4分)
由b1=
2
3,3Sn=Sn-1+2(n≥2)得3(b1+b2)=b1+2,
∴b2=
2
9,∴
b2
b1=
1
3,(5分)
∴{bn}是等比数列,公比是
1
3,∴bn=
2
3n.(6分)
(2)由(1)知cn=an•bn=
2(3n−1)
3n,
∴T
∵a5=14,a7=20,∴
a1+4d=14
a1+6d=20,解得
a1=2
d=3,
∴an=a1+(n-1)d=3n-1.(2分)
∵3Sn=Sn-1+2(n≥2)①,
∴3Sn-1=Sn-2+2(n≥3)②,
由①-②得3bn=bn-1(n≥3),
∴
bn
bn−1=
1
3(n≥3),(4分)
由b1=
2
3,3Sn=Sn-1+2(n≥2)得3(b1+b2)=b1+2,
∴b2=
2
9,∴
b2
b1=
1
3,(5分)
∴{bn}是等比数列,公比是
1
3,∴bn=
2
3n.(6分)
(2)由(1)知cn=an•bn=
2(3n−1)
3n,
∴T
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