| 1 | 1 | 1 | ||||
an=(1− | )(1− | )...[1− | ] jak chce obliczyc pierwszy wyraz tego ciagu to | |||
| 4 | 9 | (n+1)2 |
| 1 | 1 | |||
jest nim (1− | ) a drugim (1− | ) tak ![]() | ||
| 4 | 9 |
a gdyby byla suma
| 1 | ||
n =1 ⇒ ostatni czynnik iloczynu = (1− | ) = pierwszemu | |
| 4 |
| 1 | 1 | 1 | 1 | 1 | 1 | |||||||
(1− | )*(1+ | )*(1− | )*({1+ | )*(1− | )*({1+ | )*... | ||||||
| 2 | 2 | 3 | 3 | 4 | 4 |
| 1 | 1 | |||
*(1− | )*(1+ | )= | ||
| n+1 | n+1 |
| 1 | 1 | n+2 | ||||
an= | *(1+ | )= | ||||
| 2 | n+1 | 2(n+1) |