已知等差数列{an}的前n项和为Sn,且a2=3,S15=225. (1)求数列{an}的通项公式; (2)设bn=2an−2n,求数列{bn}的前n项和Tn.
问题描述:
已知等差数列{an}的前n项和为Sn,且a2=3,S15=225.
(1)求数列{an}的通项公式;
(2)设bn=2an−2n,求数列{bn}的前n项和Tn.
答
(1)设数列{an}的公差为d,依题意得:
a1+d=3 15a1+
d=22515×14 2
解得
a1=1 d=2
∴数列{an}的通项公式an=2n-1.
(2)由(1)得bn=
×4n−2n,1 2
∴Tn=b1+b2+…+bn=
(4+42+…+4n)−2(1+2+…+n)1 2
=
−n2−n
4n+1−4 6
=
×4n−n2−n−2 3
.2 3