| 5n−2 | 5 | 2 | |||
= | − | ||||
| 3n | 3 | 3n |
| 5 | 2 | 5 | 2 | ||||
− | + | − | =3 i n,k∊N+,n≠k⇔ | ||||
| 3 | 3n | 3 | 3k |
| 2 | 2 | ||
+ | =1 | ||
| n | k |
| 2k | ||
n= | , k≠2 | |
| k−2 |
| 2k−4+4 | ||
n= | ||
| k−2 |
| 2(k−2)+4 | ||
n= | ||
| k−2 |
| 4 | ||
n=2+ | ||
| k−2 |
| 13 | ||
a3= | ||
| 9 |
| 14 | ||
a6= | ||
| 9 |
| 1 | 1 | 1 | |||
+ | = | ||||
| n | k | 2 |
| 1 | ||
czyli możemy poszukać rozkładu | na sumę 2 ułamków prostych | |
| 2 |
| 1 | 3 | 2 | 1 | 1 | 1 | ||||||
= | = | + | = | + | ⇒n=3 i n=6 | ||||||
| 2 | 6 | 6 | 6 | 3 | 6 |