Respuesta :
Answer:
See below
Step-by-step explanation:
We shall prove that for all [tex]n\in\mathbb{N},3|(4^n+5)[/tex]. This tells us that 3 divides 4^n+5 with a remainder of zero.
If we let [tex]n=1[/tex], then we have [tex]4^{1}+5=9[/tex], and evidently, [tex]9|3[/tex].
Assume that [tex]4^n+5[/tex] is divisible by [tex]3[/tex] for [tex]n=k, k\in\mathbb{N}[/tex]. Then, by this assumption, [tex]3|(4^n+5)\Rightarrow4^k+5=3m,\: m\in\mathbb{Z}[/tex].
Now, let [tex]n=k+1[/tex]. Then:
[tex]4^{k+1}+5=4^k\cdot4+5\\=4^k(3+1)+5\\=3\cdot4^k+4^k+5\\=3\cdot4^k+3m\\=3(4^k+m)[/tex]
Since [tex]3|(4^k+m)[/tex], we may conclude, by the axiom of induction, that the property holds for all [tex]n\in\mathbb{N}[/tex].
