This is a classic problem I think that you mean [tex]4^{n}+6n-1[/tex]
Answer:
This problem clearly states to use proof by induction, follow the complete answer below.
Step-by-step explanation:
To prove something by induction we have to make a proposition first, in this case:
[tex]P(n): 4^{n}+6n-1[/tex] is divisible by 9
Where [tex]n \in \mathbb{N}[/tex] is the order of the proposition.
First we have to prove the first few orders, let's just check P(1):
[tex]4^1+6\times1-1=9[/tex] which is clearly divisible by 9
Now let's assume that P(n) is true that means, our original proposition is true. Now let's try to find out whether P(n+1) is true:
[tex]4^{n+1}+6(n+1)-1=4\cdot 4^n+6n+5=4\cdot n+\underbrace{4\cdot 6n-18n}_{6n}-\underbrace{4\cdot 1+9}_{5}=4(4^n+6n-1)-18n+9=4(4^n+6n-1)-9(2n-1)[/tex]
The [tex]4(4^n+6n-1)[/tex] is divisible by 9 because we assumed that "[tex](4^n+6n-1)[/tex] is divisible by 9" was a true proposition. On the other hand [tex]9(2n-1)[/tex] has a factor of 9 regardless of n, thus is also divisible by 9 that means that P(n+1) is true and it follows that P(n) must also be true, thus proving our initial statement.
