**Topic**

Prove that for all \(n \in \mathbb{N}*\), we have:

a) \({13^n} – 1\) is divisible by 6.

b) \({4^n} + 15n – 1\) is divisible by 9.

**Solution method – See details**

Prove that the statement is true for \(n \ge p\) then:

Step 1: Check the statement is true for \(n = p\)

Step 2: Suppose the proposition is true for natural numbers \(n = k \ge p\) and prove the statement true for \(n = k + 1.\) Conclusion.

**Detailed explanation**

a)

Step 1: When \(n = 1\) we have \({13^1} – 1 = 12\) divisible by 6.

So the statement is true for \(n = 1\)

Step 2: With k being an arbitrary positive integer whose proposition is true, we have to prove the proposition to be true for k + 1, that is:

\({13^{k + 1}} – 1\) is divisible by 6.

Indeed, by the assumption of induction we have:

\({13^k} – 1\) is divisible by 6.

I guess

\({13^{k + 1}} – 1 = {13.13^k} – 1 = 13.\left( {{{13}^k} – 1} \right) + 12\) divisible by 6

So the statement is true for k+1. Thus, according to the principle of mathematical induction, the statement is true for all \(n \in \mathbb{N}*\).

b)

Step 1: When \(n = 1\) we have \({4^1} + 15.1 – 1 = 18\) divisible by 9.

So the statement is true for \(n = 1\)

Step 2: With k being an arbitrary positive integer whose proposition is true, we have to prove the proposition to be true for k + 1, that is:

\({4^{k + 1}} + 15.(k + 1) – 1\) is divisible by 9.

Indeed, by the assumption of induction we have:

\({4^k} + 15k – 1\) is divisible by 9.

I guess

\({4^{k + 1}} + 15.(k + 1) – 1 = {4.4^k} + 15k + 14 = 4\left( {{4^k} + 15k – 1} \right) – 45k + 18\) divisible by 9

So the statement is true for k+1. Thus, according to the principle of mathematical induction, the statement is true for all \(n \in \mathbb{N}*\).