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:
设S1=1+
+1 12
,S2=1+1 22
+1 22
,S3=1+1 32
+1 32
,…,Sn=1+1 42
+1 n2
.1 (n+1)2
设S=
+
S1
+…+
S2
,则S=______ (用含n的代数式表示,其中n为正整数).
Sn
答
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.
∵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
=1+n(n+1)+1 n(n+1)
=1+1 n(n+1)
-1 n
,1 n+1
∴S=1+1-
+1+1 2
-1 2
+…+1+1 3
-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
,求[n(n+1)+1]2 [n(n+1)]2
,得出一般规律.
Sn
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.