| 3n − 1 | ||
an = ( | )2n − 1 | |
| 3n + 1 |
| 2 | 2n − 1 | |||
licz limn→∞[(1 − | )3n+1]k , gdzie: k = | |||
| 3n + 1 | 3n + 1 |
| 1 | 1 | |||
lim (1 − | )an = im ((1 + | )−an)−1 = | ||
| an | −an |
| −2 | ||
limn→∞(1 + | )n = e−2 | |
| n |
czyli to będzie e−k ?
k→ 2/3
| 1 | ||
odp. e−2/3 = |  | |
| e2/3 |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
| Kliknij po więcej przykładów | |
|---|---|
| Twój nick | |