分析:原题=
+
+
+
+
+
+
,又1+2+3+…+n=
,所以原式=1÷
+1÷
…+1÷
,然后再进行巧算即可.
解答:解:
1++++++=
++…
+,
=2×[(1
-)+(
-)+…+(
-)],
=2×[1-
],
=1
.
点评:在分数巧算中,
=
-(n≠0),这个规律常用.
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