题目内容

等差数列{an}中,a1=3,公差d=2,Sn为前n项和,求
1
S1
+
1
S2
+…+
1
Sn
∵等差数列{an}的首项a1=3,公差d=2,
∴前n项和Sn=na1+
n(n-1)
2
d=3n+
n(n-1)
2
×2=n2+2n(n∈N*)

1
Sn
=
1
n2+2n
=
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
3
-
1
5
)+…+(
1
n-1
-
1
n+1
)+(
1
n
-
1
n+2
)]

=
3
4
-
2n+3
2(n+1)(n+2)
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