题目内容
S=
+
+
+…+
+
.
22+1 |
22-1 |
32+1 |
32-1 |
42+1 |
42-1 |
n2+1 |
n2-1 |
(n+1)2+1 |
(n+1)2-1 |
考点:有理数无理数的概念与运算
专题:
分析:首先将原式变形为:(1+
)+(1+
)+(1+
)+…+(1+
),继而可得原式=n+1-
+
-
+
-
+…+
-
,则可求得答案.
2 |
22-1 |
2 |
32-1 |
2 |
42-1 |
2 |
(n+1)2-1 |
1 |
3 |
1 |
2 |
1 |
4 |
1 |
3 |
1 |
5 |
1 |
n |
1 |
n+2 |
解答:解:原式=(1+
)+(1+
)+(1+
)+…+(1+
)
=n+
+
+
+…+
=n+1-
+
-
+
-
+…+
-
=n+1+
-
-
=n+
-
-
.
2 |
22-1 |
2 |
32-1 |
2 |
42-1 |
2 |
(n+1)2-1 |
=n+
2 |
1×3 |
2 |
2×4 |
2 |
3×5 |
2 |
n(n+2) |
=n+1-
1 |
3 |
1 |
2 |
1 |
4 |
1 |
3 |
1 |
5 |
1 |
n |
1 |
n+2 |
=n+1+
1 |
2 |
1 |
n+1 |
1 |
n+2 |
=n+
3 |
2 |
1 |
n+1 |
1 |
n+2 |
点评:此题考查了有理式的概念与运算.解此题的关键是将原式变形为:(1+
)+(1+
)+(1+
)+…+(1+
).
2 |
22-1 |
2 |
32-1 |
2 |
42-1 |
2 |
(n+1)2-1 |
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