| 1 | ||
= lim (1 − | )(n+1)/2n =(e−1)1/2 = e−1/2 | |
| 2n |
| 1 | ||
(1− | )n+1 = | |
| 2n |
| 1 | ||
(1+ | )n+1 = | |
| −2n |
| 1 | ||
{[(1+ | )−2n]1/(−2n)}n+1= | |
| −2n |
| 1 | ||
[(1+ | )−2n](n+1)/(−2n) → e−1/2 | |
| −2n |
| 1 | ||
Bo lim ( 1 − | )2n= e−1 , więc wkładnik n + 1 trzeba podzielić przez 2n | |
| 2n |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
| Kliknij po więcej przykładów | |
|---|---|
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