题目内容
1 |
1×2 |
1 |
2 |
1 |
2×3 |
1 |
2 |
1 |
3 |
1 |
3×4 |
1 |
3 |
1 |
4 |
(1)
1 |
1×2 |
1 |
2×3 |
1 |
3×4 |
1 |
99×100 |
(2)
1 |
1×3 |
1 |
3×5 |
1 |
5×7 |
1 |
97×99 |
分析:(1)根据上述的式子总结出规律为:
=
-
,按照此规律把所求式子的每一项拆项化简,抵消合并即可求出值;
(2)根据上述的式子总结出规律为:
=
(
-
),然后按照此规律把所求式子的每一项拆项化简,抵消合并即可求出值;
1 |
n(n+1) |
1 |
n |
1 |
n+1 |
(2)根据上述的式子总结出规律为:
1 |
n(n+2) |
1 |
2 |
1 |
n |
1 |
n+2 |
解答:解:(1)
+
+
+…+
=1-
+
-
+
-
+…+
-
=1-
=
;
(2)
+
+
+…+
=
(1-
+
-
+
-
+…+
-
)
=
(1-
)
=
.
1 |
1×2 |
1 |
2×3 |
1 |
3×4 |
1 |
99×100 |
=1-
1 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
99 |
1 |
100 |
=1-
1 |
100 |
=
99 |
100 |
(2)
1 |
1×3 |
1 |
3×5 |
1 |
5×7 |
1 |
97×99 |
=
1 |
2 |
1 |
3 |
1 |
3 |
1 |
5 |
1 |
5 |
1 |
7 |
1 |
97 |
1 |
99 |
=
1 |
2 |
1 |
99 |
=
49 |
99 |
点评:此题有理数的混合运算,是一道规律型题,解答此类题常常认真观察已知的等式,进行分析,归纳总结得到一般性的规律来解决问题.根据题意归纳出一般性的规律
=
-
是解本题的关键.
1 |
n(n+1) |
1 |
n |
1 |
n+1 |

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