Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
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].
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
