Respuesta :

Answer:

The proof makes use of congruences as follows:

Step-by-step explanation:

We can prove this result using congruences module 3. First of all we shall show that

[tex]2^{2n-1}\equiv 2 \pmod{3}[/tex] for all [tex]n\in \mathbb{N}[/tex]. By induction we have

  1. [tex]n=2[/tex]. For [tex]n=2[/tex] we have [tex]2^{4-1}=8\equiv 2 \pmod{3}[/tex]
  2. Suppose that the statement is true for [tex]n=k[/tex] and let's prove that it is also true for [tex]n=k+1[/tex]. In fact,  [tex]2^{2(k+1)-1}=2^{2k-1+2}=2^{2k-1}2^{2}\equiv 2\cdot 2^{2}\equiv 8 \equiv 2 \pmod{3}[/tex]

Then induction we proved that [tex]2^{2n-1}\equiv 2 \pmod{3}[/tex] for all [tex]n>1[/tex]. Then

[tex]2^{2n-1}+1\equiv 2+1\equiv 3\equiv 0 \pmod{3}[/tex]

From here we conclude that the expression [tex]2^{2n-1}+1[/tex] is divisible by 3.