题目内容
求证：
+
+…+
=
+
+…+
．
| 1 |
| 1×2 |
| 1 |
| 3×4 |
| 1 |
| (2n-1)•2n |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| n+n |
①当n=1时，左边=
=
，右边=
=
，等式成立．
②假设当n=k时等式成立，即
+
+…+
=
+
+…+
．
则当n=k+1时，
+
+…+
+
=
+
+…+
+
=
+
+…+
+（
+
）
=
+
+…+
+（
+
-
）
=
+
+…+
+
+
=
+
+…+
+
，
即当n=k+1时，等式成立．
根据（1）（2）可知，对一切n∈N*，原等式成立．
| 1 |
| 1×2 |
| 1 |
| 2 |
| 1 |
| 1+1 |
| 1 |
| 2 |
②假设当n=k时等式成立，即
| 1 |
| 1×2 |
| 1 |
| 3×4 |
| 1 |
| (2k-1)•2k |
| 1 |
| k+1 |
| 1 |
| k+2 |
| 1 |
| 2k |
则当n=k+1时，
| 1 |
| 1×2 |
| 1 |
| 3×4 |
| 1 |
| (2k-1)•2k |
| 1 |
| (2k+1)(2k+2) |
=
| 1 |
| k+1 |
| 1 |
| k+2 |
| 1 |
| 2k |
| 1 |
| (2k+1)(2k+2) |
=
| 1 |
| k+2 |
| 1 |
| k+3 |
| 1 |
| 2k |
| 1 |
| k+1 |
| 1 |
| (2k+1)(2k+2) |
=
| 1 |
| k+2 |
| 1 |
| k+3 |
| 1 |
| 2k |
| 2 |
| 2k+2 |
| 1 |
| 2k+1 |
| 1 |
| 2k+2 |
=
| 1 |
| k+2 |
| 1 |
| k+3 |
| 1 |
| 2k |
| 1 |
| 2k+1 |
| 1 |
| 2k+2 |
=
| 1 |
| (k+1)+1 |
| 1 |
| (k+1)+2 |
| 1 |
| (k+1)+k |
| 1 |
| (k+1)+(k+1) |
即当n=k+1时，等式成立．
根据（1）（2）可知，对一切n∈N*，原等式成立．
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