题目内容
计算:
+
+
+…+
.
1 |
1×3 |
1 |
3×5 |
1 |
5×7 |
1 |
99×101 |
分析:观察原式的各项发现
=
(
-
),利用此公式对各项进行变形,然后提取
,合并抵消后即可求出值.
1 |
n(n+2) |
1 |
2 |
1 |
n |
1 |
n+2 |
1 |
2 |
解答:解:∵
=
(
-
),
∴原式=
(
-
)+
(
-
)+
(
-
)+…+
(
-
)
=
(1-
+
-
+
-
+…+
-
)
=
(1-
)
=
.
1 |
n(n+2) |
1 |
2 |
1 |
n |
1 |
n+2 |
∴原式=
1 |
2 |
1 |
1 |
1 |
3 |
1 |
2 |
1 |
3 |
1 |
5 |
1 |
2 |
1 |
5 |
1 |
7 |
1 |
2 |
1 |
99 |
1 |
101 |
=
1 |
2 |
1 |
3 |
1 |
3 |
1 |
5 |
1 |
5 |
1 |
7 |
1 |
99 |
1 |
101 |
=
1 |
2 |
1 |
101 |
=
50 |
101 |
点评:此题考查了有理数的混合运算,利用的方法是裂项相消法,培养了学生的数感、符号感,灵活运用
=
(
-
)是解本题的关键.
1 |
n(n+2) |
1 |
2 |
1 |
n |
1 |
n+2 |
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