Question

Rezolvați exercițiul din imagine

Answer 1

In avem 72 de termeni. Ii vom grupa 2 cate 2

$C=1+2^1+2^2+...+2^{71}\\ \\ C=(1+2^2)+(2^1+2^3)+(2^4+2^6)+(2^5+2^7)+...+(2^{68}+2^{70})+(2^{69}+2^{71})\\ \\ C=(1+2^2)+2(1+2^2)+2^4(1+2^2)+2^5(1+2^2)+...+2^{68}(1+2^2)+2^{69}(1+2^2)\\ \\C=5+2\cdot5+2^4\cdot5+2^5\cdot5+...+2^{68}\cdot5+2^{69}\cdot5\\ \\ C=5(1+2+2^4+2^5+...+2^{68}+2^{69}) \Rightarrow C~divizibil~cu~5$

Answer 2

C=1+2¹+2²+....+2⁷¹=

(1+2²)+(2+2³)+(2⁴+2⁶)+(2⁵+2⁷)+...+(2⁶⁸+2⁷⁰)+(2⁶⁹+2⁷¹)=

1+2²+2×(1+2²)+2⁴×(1+2²)+2⁵×(1+2²)+...+2⁶⁸×(1+2²)+2⁶⁹×(1+2²)=

5+2×5+2⁴×5+2⁵×5+....+2⁶⁸×5+2⁶⁹×5=

5+(1+2+2⁴+2⁵+2⁸+....+2⁶⁸+2⁶⁹) => ⋮ 5