已知数列{an}中,a1=2,其前n项和Sn满足Sn+1-Sn=2n+1(n∈N+).(1)求数列{an}的通项公式an以及前n和Sn

2025-05-17 13:19:11
推荐回答(1个)
回答1:

(1)∵Sn+1-Sn=2n+1
∴an+1=2n+1
∴an=2n
∴Sn=

2(1?2n)
1?2
=2n+1-2;
(2)bn=2log2an+1=2n+1,
1
bn?bn+1
=
1
2
1
2n+1
-
1
2n+3
),
∴Tn=
1
2
1
3
-
1
5
+
1
5
-
1
7
+…+
1
2n+1
-
1
2n+3
)=
1
2
1
3
-
1
2n+3
)=
n
6n+9