设数列{an}是一等差数列,数列{bn}的前n项和为Sn=23(bn−1),若a2=b1,a5=b2.
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设数列{an}是一等差数列,数列{bn}的前n项和为S
(1)∵S1=
2
3(b1−1)=b1,∴b1=-2,
又S2=
2
3(b2−1)=b1+b2=−2+b2,∴b2=4,∴a2=-2,a5=4,(2分)
∵an为一等差数列,∴公差d=
a5−a2
3=
6
3=2,(4分)
即an=-2+(n-2)•2=2n-6.(6分)
(2)∵Sn+1=
2
3(bn+1−1)①,Sn=
2
3(bn−1)②,
①-②得Sn+1−Sn=
2
3(bn+1−bn)=bn+1,∴bn+1=-2bn,(9分)
∴数列bn是一等比数列,公比q=-2,b1=-2,即bn=(-2)n.
∴Sn=
2
3[(−2)n−1].(12分)
2
3(b1−1)=b1,∴b1=-2,
又S2=
2
3(b2−1)=b1+b2=−2+b2,∴b2=4,∴a2=-2,a5=4,(2分)
∵an为一等差数列,∴公差d=
a5−a2
3=
6
3=2,(4分)
即an=-2+(n-2)•2=2n-6.(6分)
(2)∵Sn+1=
2
3(bn+1−1)①,Sn=
2
3(bn−1)②,
①-②得Sn+1−Sn=
2
3(bn+1−bn)=bn+1,∴bn+1=-2bn,(9分)
∴数列bn是一等比数列,公比q=-2,b1=-2,即bn=(-2)n.
∴Sn=
2
3[(−2)n−1].(12分)
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