Wzór rekurencyjny
Mateusz: Wyznacz kolejne wyrazy ciągu (a1−a5) opisanego wzorem rekurencyjnym:
a1=1/4 , an+1=(an)2*4n
5 maj 18:39
Janek191:
a
n+1 = ( a
n)
2 * 4
n
więc
| | 1 | | 1 | | 1 | |
a2 = ( a1)2 * 41 = ( |
| )2 *4 = |
| *4 = |
| |
| | 4 | | 16 | | 4 | |
| | 1 | | 1 | |
a3 = ( a2)2 * 42 = ( |
| )2 *16 = |
| *16 = 1 |
| | 4 | | 16 | |
a
4 = ( a
3)
2 *4
3 = 1
2 *64 = 1*64 = 64
a
5 = ( a
4)
2* 4
4 = 64
2 *256 = 4 096*256 = 1 048 576
6 maj 07:33