PROSZĘ O POMOC ;<
Banius: wykaz ze liczby 8/√5+2 , √5−1/√5+1 ,√5−1/16 tworza ciag geometryczny.
20 lut 17:01
M:
5 sty 06:06
Jolanta: W ciągu geometrycznym a
n2=a
n−1*a
n+1. a
22=a
1*a
3
| | √5−1 | | 8 | | √5−1 | |
( |
| )2= |
| * |
| |
| | √5+1 | | √5+2 | | 16 | |
| 5−2√5+1 | | √5−1 | |
| = |
| |
| 5+2√5+1 | | 2*(√5+2) | |
| 2(3−√5 | | (√5−1)(√5−2) | |
| = |
| |
| 2(3+√5 | | 2(√5+2)(√5−2) | |
| (3−√5)2 | | 7−3√5 | |
| = |
| |
| (3+√5)(3−√5 | | ,2 | |
a
22=a
1*a
3. ciag jest geometryczny
5 sty 20:23