Question

va rog ajutatima repede

Answer 1

19. (2^n+2+2^n)/(4^n+1+4^n)=
=(2^n+2+2^n)/(2^2n+2+2^2n)=
=2^n(2²+2⁰)/[2^2n(2²+2⁰)]=
=2^n × 5/2^2n ×5=
=2^n/2^2n=
=2/2²=
=1/2

20.5^(n+1) + 4*5^n
5^n(5+4)=5^n * 9
10^n+1-10^n
10^n(10¹-10⁰)
2^n * 5^n(10-1)
2^n * 5^n *9 divizibil cu 5^n*9


21. n ∈ N*⇒1^n =1 - impar
⇒ 2^n=2- par
⇒3^n=3 - impar
1^n+2^n+3^n = ( impar )+(par )+(impar)=(par)

22. Cazul I: Fie n=4k+1
1^n = 1 indiferent de n
2^(4k+1)= 2^1 =2
3^(4k+1)= 3^1 = 3
4^(4k+1) = 4^1 =4
Cazul II : Fie n= 4k+2
2^(4k+2) = 2^2 = 4
3^(4k+2) =3^2 = 9
4^(4k+2) = 6
Cazul III Fie n=4k+3
2^(4k+3) = 2^3 =8
3^(4k+3) = 3^3 = 7
4^(4k+3) = 4