podaj wzór ogólny ciągu rekurencyjnego
iwonaaa: | | ⎧ | 2 dla n=1 | |
| |an| = | ⎨ | |
|
| | ⎩ | 2an−1 −1 dla n ≥ 2 | |
16 maj 10:46
Adamm: |an| ?
16 maj 10:49
kochanus_niepospolitus:
a2 = 2*2 − 1 = 3 = 21 + 1
a3 = 2*3 − 1 = 5 = 22 + 1
a4 = 2*5 − 1 = 9 = 23 + 1
a5 = 2*9 − 1 = 17 = 24 + 1
a6 = 2*(24+1) − 1 = 25 + 2 − 1 = 25 + 1
itd.
an = 2n−1 + 1
16 maj 10:52
Adamm: A(x)=∑
n=1∞a
nx
n−1=a
1+∑
n=2∞a
nx
n−1=
=a
1+∑
n=1∞a
n+1x
n=a
1+x∑
n=1∞(2a
n−1)x
n−1=
| | 2−3x | | A | | B | |
A(x)= |
| = |
| + |
| |
| | (1−x)(1−2x) | | 1−x | | 1−2x | |
2−3x=A−2Ax+B−Bx
A+B=2, 2A+B=3
A=1, B=1
| | 1 | | 1 | |
A(x)= |
| + |
| =∑n=1∞xn−1+∑n=1∞2n−1xn−1= |
| | 1−x | | 1−2x | |
=∑
n=1∞(2
n−1+1)x
n−1
a
n=2
n−1+1
16 maj 11:02