Question

Demonstrati ca (3 + 3^{3} + 3^{5} + ... + 3^{19} | 30 .
| - divide .$x^{123}$

Answer 1

Grupam termenii cate 2=>(3+27)+3 la puterea4(3+27)+....3la puterea16(3+27)=30(1+3 la puterea4+...3 la puterea16)=>. deoarece unuk din factori este30=> nr este multiplu de 30

Answer 2

 [tex]Avem:\;\\
.\;\;\;3^1+3^3+3^5+3^7+3^9+3^{11}+3^{13}+3^{15}+3^{17}+3^{19}=\\
=(\underbrace{3+27}_{=30})+3^4(3+27)+3^8(3+27)+3^{12}(3+27)+3^{16}(3+27)=\\
=30\cdot(1+3^4+3^8+3^{12}+3^{16})\,\vdots\,30 [/tex]