obl granice ciagu
ola: an = (1 + 5n+3)2n+7
14 maj 10:39
Basia:
2n+7 = n+3 + n+3 + 1 = 2(n+3)+1
(1+
5n+3)
2n+7 = (1+
5n+3)
2(n+3)+1 =
[(1+
5n+3)
n+3]
2*(1+
5n+3)
1 =
| | 1 | |
[(1+ |
| )n+35*5]2*(1+5n+3) = |
| | n+35 | |
| | 1 | |
[(1+ |
| )n+35]10*(1+5n+3) → e10*1 = e10 |
| | n+35 | |
15 maj 00:06