**Topic**

Prove that \({a^n} – {b^n} = (a – b)({a^{n – 1}} + {a^{n – 2}}b + … + a{b^{) n – 2}} + {b^{n – 1}})\) for all \(n \in \mathbb{N}*\)

**Solution method – See details**

Inductive method: Prove that the statement is true with \(n \ge p\)

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**

Step 1: When \(n = 1\) we have \({a^1} – {b^1} = a – b\) which is obviously true

So the equality holds for \(n = 1\)

Step 2: With k being an arbitrary positive integer whose equality is true, we must prove the equality true for k + 1, ie:

\({a^{k + 1}} – {b^{k + 1}} = (a – b)({a^{k + 1 – 1}} + {a^{k + 1 – 2} }b + … + a{b^{k + 1 – 2}} + {b^{k + 1 – 1}})\) or \({a^{k + 1}} – {b^{k + 1}} = (a – b)({a^k} + {a^{k – 1}}b + … + a{b^{k – 1}} + {b^k})\)

Indeed, by the assumption of induction we have:

\({a^k} – {b^k} = (a – b)({a^{k – 1}} + {a^{k – 2}}b + … + a{b^{k –) 2}} + {b^{k – 1}})\)

I guess

\(\begin{array}{l}{a^{k + 1}} – {b^{k + 1}} = a. {a^k} – b. {b^k} = a\left( {{a^k} – {b^k}} \right) + a{b^k} – b. {b^k} = a\left( {{a^k} – {b^k}} \ right) + \left( {a – b} \right). {b^k}\\ = a.(a – b)({a^{k – 1}} + {a^{k – 2}} b + … + a{b^{k – 2}} + {b^{k – 1}}) + \left( {a – b} \right) {b^k}\\ = (a – b )\left[ {a({a^{k – 1}} + {a^{k – 2}}b + … + a{b^{k – 2}} + {b^{k – 1}}) + {b^k}} \right]\\ = (a – b)({a^k} + {a^{k – 1}}b + … + a{b^{k – 1}} + {b^k})\end{array} \)

So the equality holds for k+1. Thus, by the principle of mathematical induction, the equality holds for all \(n \in \mathbb{N}*\).