Question

Comparati numerele:A=2^2n+5*3^n+5*4^n si B=6^n+1*8^n+4,unde n este numar natural.

Answer 1

A=2^(2n+5)* 3^(n+5) *4^n=2^(2n+5) *3^(n+5) *2^(2n)=
2^(4n+5) *3^(n+5)=(factor comun)=(2^(4n+5) *3^(n+1))*(1*3^4)

A=(3^4)*(2^(4n+5) *3^(n+1))=(3^4)*FC, FC=factorul comun

B=6^(n+1) *8^(n+4)=2^(n+1) *3^(n+1) *(2^3)^(n+4)=
2^(n+1) *3^(n+1) *2^(3n+12)=2^(4n+13) *3^(n+1)=(acelasi factor comun ca la ''A'')=(2^(4n+5) *3^(n+1))*(2^8 *1)=(2^(4n+5) *3^(n+1))*2^8=
(2^(4n+5) *3^(n+1))*(2^2)^4

B=(4^4)*(2^(4n+5) *3^(n+1))=(4^4)*FC

Se observa ca A<B