| a3−b3 | ||
a−b= | ||
| a2+ab+b2 |
| 1+3+5+...+(2n−1 | ||
A co będzie w tym: an= | −n | |
| n+1 |
| 1+(2n−1) | 1 | |||
an= | *n*(n+1)−n=n3+n2−n= | →∞ Tak? | ||
| 2 | 0 |
| 1+2n−1 | 1 | n2 | n(n+1) | ||||
*n* | −n= | − | = | ||||
| 2 | n+1 | n+1 | n+1 |
| n | |
→1 | |
| n+1 |
| −n | |
→−1 | |
| n+1 |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
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