Question

Prove that 9 divides 4" + 6n – 1 for n >1 by mathematical induction.

Answer 1

This is a classic problem I think that you mean $4^{n}+6n-1$

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:

$P(n): 4^{n}+6n-1$ is divisible by 9

Where $n \in \mathbb{N}$ is the order of the proposition.

First we have to prove the first few orders, let's just check P(1):

$4^1+6\times1-1=9$ 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:

$4^{n+1}+6(n+1)-1=4\cdot 4^n+6n+5=4\cdot n+\underbrace{4\cdot 6n-18n}_{6n}-\underbrace{4\cdot 1+9}_{5}=4(4^n+6n-1)-18n+9=4(4^n+6n-1)-9(2n-1)$

The $4(4^n+6n-1)$ is divisible by 9 because we assumed that "$(4^n+6n-1)$ is divisible by 9" was a true proposition. On the other hand $9(2n-1)$ 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.