## Solving Lesson 7 Page 29 Math Learning Topic 10 – Kite>

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 $$n \in \mathbb{N}*$$.