题目
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| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2n |
答案
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2k+1 |
当n=k时,f(k+1)=1+
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| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2 k |
则f(k+1)-f(k)=1+
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| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2 k |
| 1 |
| 2 k+1 |
| 1 |
| 2k+1 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2 k |
=
| 1 |
| 2k+1 |
| 1 |
| 2k+2 |
| 1 |
| 2k+1 |
故答案为:
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| 2k+1 |
| 1 |
| 2k+2 |
| 1 |
| 2k+1 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2n |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2k+1 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2 k |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2 k |
| 1 |
| 2 k+1 |
| 1 |
| 2k+1 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2 k |
| 1 |
| 2k+1 |
| 1 |
| 2k+2 |
| 1 |
| 2k+1 |
| 1 |
| 2k+1 |
| 1 |
| 2k+2 |
| 1 |
| 2k+1 |