Question

设{an}是等差数列，其前n项和为Sn，已知S7=63，a4+a5+a6=33，...

设{an}是等差数列，其前n项和为Sn，已知S7=63，a4+a5+a6=33， (1)写出数列{an}的通项公式; (2) 求数列bn=2an+n，求数列{bn}的前n项和Tn; (3) 求证:1S1+1S2+1S3+…+1Sn＜34

Answer 1

解答:解:(1)∵s7=
(a1+a7)
2
×7=7a4=63
∴a4=9，又a4+a5+a6=33，3a5=33，则a5=11
公差d=2，an=2n+1;
(2)∵bn=2an+n=22n+1+n
∴Tn=b1+b2+…+bn=(23+1)+(25+2)+••+(22n+1+n)
=(23+25+…+22n+1)+(1+2+…+n)
=
8(4n-1)
3
+
n(n+1)
2
(3)由等差数列的前n项和公式可得，Sn=3n+
n(n-1)
2
×2=n2+2n=n(n+2)
∴
1
Sn
=
1
n(n+2)
=
1
2
(
1
n
-
1
n+2
)
∴
1
S1
+
1
S2
+…+
1
Sn
=
1
2
(1-
1
3
+
1
2
-
1
4
+…+
1
n
-
1
n+2
)
=
1
2
(1+
1
2
-
1
1+n
-
1
n+2
)=
3
4
-
2n+3
2(n+1)(n+2)
＜
3
4