| 2n+2 | ||
wykazać , że ciag którego n−ta suma częściowa Sn wyraża się wzorem Sn=12− | jest | |
| 3n−1 |
| 2n+2 | ||
Sn=12− | ||
| 3n−1 |
| 2n | ||
zatem an=− | *30 | |
| 3n |
| 8 * 2n−1 | 2 | ||
= 8 * | n−1 | ||
| 3n−1 | 3 |
| 2 | 16 | 2 | ||||
Sn = 12 − 8 * | n−1 = 12 − | * | n−2 | |||
| 3 | 3 | 3 |
| 2 | ||
Sn−1 = 12 − 8 * | n−2 | |
| 3 |
| 16 | 2 | 2 | 2 | |||||
an =Sn − Sn−1 = 12 − | * | n−2 − 12 + 8 * | n−2 = | n−2( − | ||||
| 3 | 3 | 3 | 3 |
| 16 | 2 | 8 | 2 | 2 | 8 | |||||||
+8) = | n−2 * | = | n−1 * | * | ||||||||
| 3 | 3 | 3 | 3 | 3 | 3 |
| 16 | 2 | |||
=12− | * | n−2 dlaczego zrobiło się n−2? jak? | ||
| 3 | 3 |
| 2 | 2 | 2 | 2 | ||||
* | n−2 = | n−2+1 = | n−1 | ||||
| 3 | 3 | 3 | 3 |
| 2 | 24 | 2 | ||||
=12−8* | n−1=12− | * | n−1 | |||
| 3 | 3 | 3 |