已知等差数列{an}的前n项和为Sn,且a3=5,S15=150.(1)求数列{an}的通项公式;(2)设bn=2an+(-1)nan

2025-05-11 17:02:56
推荐回答(1个)
回答(1):

(1)∵等差数列{an}的前n项和为Sn,且a3=5,S15=150,

15a1+
15×14
2
d=150
a1+2d=5

解得a1=3,d=1,
∴an=3+(n-1)=n+2.
(2)bn=2an+(-1)nan=2n+2+(-1)n?(n+2),
当n为偶数时,
Tn=23-3+24+4+…+2n+2+(-1)n?(n+2)
=(23+24+…+2n+2)+(-3+4)+…+(-n-1+n+2)
=
8(1?2n)
1?2
+
n
2

=2n+3+
n
2
?8

当n为奇数时,
Tn=23-3+24+4+…+2n+2+(-1)n?(n+2)
=(23+24+…+2n+2)+(-3+4)+…+(-n+n+1)-(n+2)
=
8(1?2n)
1?2
+
n?1
2
?n?2

=2n+3?
n
2
?
21
2

∴Tn=
2n+3+
n
2
?8,n为正偶数
2n+3?
n
2
?
21
2
,n为正奇数