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观察下面的式子:
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

(1)计算:
S1
=
3
2
3
2
S3
=
13
12
13
12
;猜想
Sn
=
n(n+1)+1
n(n+1)
n(n+1)+1
n(n+1)
(用n的代数式表示);
(2)计算:S=
S1
+
S2
+
S3
+…+
Sn
(用n的代数式表示).
分析:(1)分别求出S1,S2,…的值,再求出其算术平方根即可;
(2)根据(1)的结果进行拆项得出1+
1
2
+1+
1
6
+1+
1
12
+…+1+
1
n(n+1)
,再转换成n+(1-
1
2
+
1
2
-
1
3
+
1
3
-
1
4
+…+
1
n
-
1
n+1

即可求出答案.
解答:(1)解:∵S1=1+
1
12
+
1
22
=
9
4

S1
=
9
4
=
3
2

∵S2=1+
1
22
+
1
32
=
49
36

S2
=
7
6

∵S3=1+
1
32
+
1
42
=
169
144

S3
=
13
12

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

Sn
=
n2+n+1
n(n+1)
=
n(n+1)+1
n(n+1)

故答案为:
3
2
13
12
n(n+1)+1
n(n+1)


(2)解:S=
3
2
+
7
6
+
13
12
+…+
n(n+1)+1
n(n+1)

=1+
1
2
+1+
1
6
+1+
1
12
+…+1+
1
n(n+1)

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

=n+1-
1
n+1

=
n2+2n
n+1
点评:本题考查了二次根式的化简,主要考学生的计算能力,题目比较好,但有一定的难度.
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