设S1=1+112+122,S2=1+122+132,S3=1+132+142,…,Sn=1+1n2+1(n+1)2.设S=S1+S2+…+Sn,则S=______ (用含n的代数式表示,其中n为正整数).

问题描述:

S1=1+

1
12
+
1
22
S2=1+
1
22
+
1
32
S3=1+
1
32
+
1
42
,…,Sn=1+
1
n2
+
1
(n+1)2

S=
S1
+
S2
+…+
Sn
,则S=______ (用含n的代数式表示,其中n为正整数).

∵Sn=1+

1
n2
+
1
(n+1)2
=
n2(n+1)2+(n+1)2+n2 
n2(n+1)2
=
[n(n+1)]2+2n2+2n+1
[n(n+1)]2
=
[n(n+1)+1]2
[n(n+1)]2

Sn
=
n(n+1)+1
n(n+1)
=1+
1
n(n+1)
=1+
1
n
-
1
n+1

∴S=1+1-
1
2
+1+
1
2
-
1
3
+…+1+
1
n
-
1
n+1

=n+1-
1
n+1

=
(n+1)2−1
n+1
=
n2+2n
n+1

故答案为:
n2+2n
n+1

答案解析:由Sn=1+
1
n2
+
1
(n+1)2
=
n2(n+1)2+(n+1)2+n2 
n2(n+1)2
=
[n(n+1)]2+2n2+2n+1
[n(n+1)]2
=
[n(n+1)+1]2
[n(n+1)]2
,求
Sn
,得出一般规律.
考试点:二次根式的化简求值.

知识点:本题考查了二次根式的化简求值.关键是由Sn变形,得出一般规律,寻找抵消规律.