## Solving Lesson 8 Page 38 Math Learning Topic 10 – Kite>

Topic

Prove Newton’s binomial formula by induction:

$${(a + b)^n} = C_n^0{a^n} + C_n^1{a^{n – 1}}b + … + C_n^{n – 1}a{b^{n – 1}} + C_n^n{b^n}$$ with $$n \in \mathbb{N}*$$

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

Newton’s binomial formula: $${(a + b)^n} = C_n^0{a^n} + C_n^1{a^{n – 1}}b + … + C_n^{n – 1} a{b^{n – 1}} + C_n^n{b^n}$$

We prove Newton’s binomial formula by induction on n.

Step 1: With $$n = 1$$ we have $${(a + b)^1} = C_1^0a + C_1^1b\quad ( = a + b)$$

So the formula is true for $$n = 1$$

Step 2: Assuming the formula is true for $$n = k$$, that is, yes:

$${(a + b)^k} = C_k^0{a^k} + C_k^1{a^{k – 1}}b + … + C_k^{k – 1}a{b^{k) – 1}} + C_k^k{b^k}$$

We will prove that the formula is also true for $$n = k + 1$$, that is, we need to prove

$${(a + b)^{k + 1}} = C_{k + 1}^0{a^{k + 1}} + C_{k + 1}^1{a^k}b + … + C_{k + 1}^ka{b^k} + C_{k + 1}^{k + 1}{b^{k + 1}}$$

Indeed we have

$$\begin{array}{l}{(a + b)^{k + 1}} = {(a + b)^k}(a + b) = \left( {C_k^0{a^k) } + C_k^1{a^{k – 1}}b + … + C_k^{k – 1}a{b^{k – 1}} + C_k^k{b^k}} \right)(a + b)\\ = \left( {C_k^0{a^k} + C_k^1{a^{k – 1}}b + … + C_k^{k – 1}a{b^{k – 1 }} + C_k^k{b^k}} \right)a + \left( {C_k^0{a^k} + C_k^1{a^{k – 1}}b + … + C_k^{k – 1}a{b^{k – 1}} + C_k^k{b^k}} \right)b\\ = \left( {C_k^0{a^{k + 1}} + C_k^1 {a^k}b + … + C_k^{k – 1}{a^2}{b^{k – 1}} + C_k^ka{b^k}} \right) + \left( {C_k^ 0{a^k}b + C_k^1{a^{k – 1}}{b^2} + … + C_k^{k – 1}a{b^k} + C_k^k{b^{k + 1}}} \right)\\ = C_k^0{a^{k + 1}} + \left( {C_k^1 + C_k^0} \right){a^k}b + … + \left ( {C_k^m + C_k^{m – 1}} \right){a^{k + 1 – m}}{b^m} + … + \left( {C_k^k + C_k^{k – 1 }} \right)a{b^k} + C_k^k{b^{k + 1}}\end{array}$$

Which $$C_k^m + C_k^{m – 1} = C_{k + 1}^m\;(0 \le m \le k),\;C_k^0 = C_{k + 1}^0 = 1,C_k^k = C_{k + 1}^{k + 1} = 1$$

$$\Rightarrow {(a + b)^{k + 1}} = C_{k + 1}^0{a^{k + 1}} + C_{k + 1}^1{a^k}b + … + C_{k + 1}^ka{b^k} + C_{k + 1}^{k + 1}{b^{k + 1}}$$

So the formula is true for all natural numbers $$n \ge 1$$