postać rekurencyjna
Bartek: Znajdz postać rekurencyjną ciągu:
2n + 5n
12 cze 10:24
Mariusz:
∑
n=0∞2
nx
n+∑
n=0∞5
nx
n
| 1 | | 1 | | 1−5x+1−2x | |
| + |
| = |
| |
| 1−2x | | 1−5x | | (1−2x)(1−5x) | |
A(x)(1−7x+10x
2)=−7x+2
(∑
n=0∞(a
nx
n))(1−7x+10x
2)=−7x+2
a
0+a
1x+∑
n=2∞(a
nx
n)−7x((∑
n=0∞(a
nx
n)))+
10x
2((∑
n=0∞(a
nx
n)))=−7x+2
a
0+a
1x+∑
n=2∞(a
nx
n)−7x((∑
n=1∞(a
n−1x
n−1)))+
10x
2((∑
n=2∞(a
n−2x
n−2)))=−7x+2
a
0+a
1x+∑
n=2∞(a
nx
n)+((∑
n=1∞(−7a
n−1x
n)))+
((∑
n=2∞(10a
n−2x
n)))=−7x+2
a
0+a
1x+∑
n=2∞(a
nx
n)+((∑
n=2∞(−7a
n−1x
n))−7a
0x)+
((∑
n=2∞(10a
n−2x
n)))=−7x+2
a
0+a
1x−7a
0x=−7x+2
a
0=2
a
1−7a
0=−7
a
0=2
a
1=−7+7a
0
a
0=2
a
1=7
a
0=2
a
1=7
a
n−7a
n−1+10a
n−2=0
a
0=2
a
1=7
a
n=7a
n−1−10a
n−2
12 cze 11:32