当n≥2时,bn=an+1−an=
an−1+an |
2 |
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1 |
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所以{bn}是以1为首项,−
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(2)解由(1)知bn=an+1−an=(−
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当n≥2时,an=a1+(a2-a1)+(a3-a2)++(an-an-1)=1+1+(-
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1 |
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=1+
1−(−
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1−(−
|
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1 |
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5 |
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3 |
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当n=1时,
5 |
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所以an=
5 |
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an+an+1 |
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an−1+an |
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1 |
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1 |
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1 |
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1 |
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1 |
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1−(−
| ||
1−(−
|
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1 |
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5 |
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3 |
1 |
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5 |
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3 |
1 |
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5 |
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1 |
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