已知数列{an}的前n项和为Sn,且Sn=2an-2,数列{bn}满足b1=1,且bn+1=bn+2.(1)求数列{an},{bn}的通项公式;(2)设cn=1-(-1)n2an-1+(-1)n2bn,求数列{cn}的前2n项和T2n.
问题描述:
已知数列{an}的前n项和为Sn,且Sn=2an-2,数列{bn}满足b1=1,且bn+1=bn+2.
(1)求数列{an},{bn}的通项公式;
(2)设cn=
an-1-(-1)n
2
bn,求数列{cn}的前2n项和T2n.1+(-1)n
2
答
(1)当n=1,a1=2; …(1分)当n≥2时,an=Sn-Sn-1=2a...