√4pn2−1 − √4pn2+1 | ||
limn−>inf( | + | |
√6n2+1 − √6n2−1 |
(1−√3pn4−2√3pn6)2 | |
) = −2 | |
(3√n−√6pn4−p3)3) |
3 | ||
−√ | ||
2p |
−12p | |
√216p3 |
3 | 12p | |||
−√ | − | = −2 | ||
2p | √216p3 |
25 | ||
z tego wychodzi ze p = | ||
24 |
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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Twój nick | |