pn2−1 | ||
bn= | ||
(p−1)n2+1 |
p | |
=p | |
p−1 |
n2(p−1/n2 | p | |||
Jesli p−1≠0 to limn→∞bn= | = | |||
n2(p−1)+1/n2 | p−1 |
p | |
=p | |
p−1 |
−1 | ||
bn= | ||
−n2+1 |
2n2−1 | ||
bn= | ||
n2+1 |
4−2n | ||
mam ciag an= | ||
3n+5 |
2n−4 | ||
Moge go zapisac tak an= − | ? | |
3n+5 |
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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