| an | ||
Wiedząc, że a1=2 oraz an+1= | dla n≥1 wykaż, że ciąg(bn) jest ciągiem | |
| 2+an |
| an | ||
geometrycznym jeśli bn= | ||
| 1−an |
| a1 | 2 | |||
b1 = | = | = −2 | ||
| 1−a1 | 1−2 |
| an+1 | ||
bn+1 = | = | |
| 1−an+1 |
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= | ||||||||||
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= | |||||||
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| an | 2+an | an | |||
* | = | ||||
| 2+an | 2 | 2 |
| an−1 | ||
z tego wynika, że b1 = −2 i dla n≥2 bn = | ||
| 2 |
| a1 | ||
czyli b2 = | = 1 | |
| 2 |
| a2 | ||
b3 = | ||
| 2 |
| 2 | 1 | |||
a2 = | = | |||
| 2+2 | 2 |
| 1 | ||
b3 = | ||
| 4 |