Question

Compara numerele
A=2(1+3+3^2+....+3^20)
B=(3^7)^5•3^10:9^12
Ajutor

Answer 1

$A=2(1+3+3^2+....+3^{20})=2(3^{0}+3^{1+2+3+...+20} )=2+2\cdot3 ^{210} \\ \\ 1+2+3+....+20=(20\cdot21):2=210 \\ \\ B=(3^{7})^{5} \cdot3^{10} :9^{12} = 3^{35}\cdot3^{10}:3^{2\cdot12} =3^{35+10}:3^{24} =3^{45}:3 ^{24}=3^{45-24} \\ \\ B=3^{11} ; A= 2+2\cdot3 ^{210} \\ \\ A\ \textgreater \ B$

Answer 2

Am atasat rezolvarea!