| 1+x | ||
Sn= | *x | |
| 2 |
| 1+x | |
*x=153 | |
| 2 |
! an =x −−− wyraz ostatni , n −−− ilość wyrazów w tej sumie!
poprawiam
a1= 1 r = 4 an = x , x>0
więc an = a1+(n−1)*r => 1 (n−1)*4= x => 4n =x+3
| x+3 | ||
to: n = | .... i n€N+
| |
| 4 |
| 1+x | x+3 | |||
Sn=153 => | * | = 153 /*8
| ||
| 2 | 4 |
| 33+3 | ||
zatem ilość wyrazów w tej sumie jest: n = | = 9
| |
| 4 |
| 1+33 | ||
s9= | *9= 17*9 = 153 ... czyli ok. | |
| 2 |