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bn2an−2n,求数列{bn}的前n项和Tn

(1)设数列{an}的公差为d,依题意得:

a1+d=3
15a1+
15×14
2
d=225

解得
a1=1
d=2

∴数列{an}的通项公式an=2n-1.
(2)由(1)得bn
1
2
×4n−2n

∴Tn=b1+b2+…+bn=
1
2
(4+42+…+4n)−2(1+2+…+n)

=
4n+1−4
6
n2−n

=
2
3
×4nn2−n−
2
3