设数列{an}是一等差数列,数列{bn}的前n项和为Sn=2/3(bn−1),若a2=b1,a5=b2. (1)求数列{an}的通项公式; (2)求数列{bn}的前n项和Sn.
问题描述:
设数列{an}是一等差数列,数列{bn}的前n项和为Sn=
(bn−1),若a2=b1,a5=b2.2 3
(1)求数列{an}的通项公式;
(2)求数列{bn}的前n项和Sn.
答
(1)∵S1=
(b1−1)=b1,∴b1=-2,2 3
又S2=
(b2−1)=b1+b2=−2+b2,∴b2=4,∴a2=-2,a5=4,(2分)2 3
∵an为一等差数列,∴公差d=
=
a5−a2
3
=2,(4分)6 3
即an=-2+(n-2)•2=2n-6.(6分)
(2)∵Sn+1=
(bn+1−1)①,Sn=2 3
(bn−1)②,2 3
①-②得Sn+1−Sn=
(bn+1−bn)=bn+1,∴bn+1=-2bn,(9分)2 3
∴数列bn是一等比数列,公比q=-2,b1=-2,即bn=(-2)n.
∴Sn=
[(−2)n−1].(12分)2 3