Exploring the Power of the Binomial Theorem Through Mathematical Induction

Introduction

Have you ever wondered how expressions like (z1+z2)n(z_1 + z_2)^n expand so neatly into a sum of terms? Today, we’re diving into the Binomial Theorem, a powerful tool in algebra that reveals the secret behind these expansions. Our goal: to prove the formula

(z1+z2)n=∑k=0n(nk)z1n−kz2k,for n=1,2,…(z_1 + z_2)^n = \sum_{k=0}^n \binom{n}{k} z_1^{n-k} z_2^k, \hspace{3mm} \text{for } n = 1, 2, \ldots

where

(nk)=n!k!(n−k)!,k=0,1,2,…,n.\binom{n}{k} = \frac{n!}{k!(n-k)!}, \hspace{3mm} k = 0, 1, 2, \ldots, n.

Don’t worry if the notation seems overwhelming—we’ll break everything down step by step!

What Is the Binomial Theorem?

The Binomial Theorem gives a formula to expand expressions of the form (z1+z2)n(z_1 + z_2)^n into a sum of terms involving powers of z1z_1 and z2z_2. Each term in the expansion involves a binomial coefficient, (nk)\binom{n}{k}, which tells us how many ways we can choose kk objects from nn objects.

For example, if n=3n = 3, the theorem says:

(z1+z2)3=(30)z13z20+(31)z12z21+(32)z11z22+(33)z10z23.(z_1 + z_2)^3 = \binom{3}{0} z_1^3 z_2^0 + \binom{3}{1} z_1^2 z_2^1 + \binom{3}{2} z_1^1 z_2^2 + \binom{3}{3} z_1^0 z_2^3.

Simplifying the binomial coefficients, we get:

(z1+z2)3=z13+3z12z2+3z1z22+z23.(z_1 + z_2)^3 = z_1^3 + 3z_1^2z_2 + 3z_1z_2^2 + z_2^3.

Our Goal: Prove the General Formula

To prove the binomial formula works for any positive integer nn, we’ll use a method called mathematical induction.

The Proof

Step 1: Base Case

Let’s start by verifying the formula for n=1n = 1:

(z1+z2)1=z1+z2.(z_1 + z_2)^1 = z_1 + z_2.

The formula says:

∑k=01(1k)z11−kz2k=(10)z11z20+(11)z10z21.\sum_{k=0}^1 \binom{1}{k} z_1^{1-k} z_2^k = \binom{1}{0} z_1^1 z_2^0 + \binom{1}{1} z_1^0 z_2^1.

Simplify each term:

(10)z11z20=1⋅z1=z1,(11)z10z21=1⋅z2=z2.\binom{1}{0} z_1^1 z_2^0 = 1 \cdot z_1 = z_1, \hspace{5mm} \binom{1}{1} z_1^0 z_2^1 = 1 \cdot z_2 = z_2.

So, the formula holds for n=1n = 1.

Step 2: Inductive Hypothesis

Assume the formula is true for some positive integer n=mn = m:

(z1+z2)m=∑k=0m(mk)z1m−kz2k.(z_1 + z_2)^m = \sum_{k=0}^m \binom{m}{k} z_1^{m-k} z_2^k.

This is our inductive hypothesis.

Step 3: Inductive Step

We need to prove that the formula holds for n=m+1n = m + 1:

(z1+z2)m+1=∑k=0m+1(m+1k)z1m+1−kz2k.(z_1 + z_2)^{m+1} = \sum_{k=0}^{m+1} \binom{m+1}{k} z_1^{m+1-k} z_2^k.

Using the distributive property, expand (z1+z2)m+1(z_1 + z_2)^{m+1}:

(z1+z2)m+1=(z1+z2)⋅(z1+z2)m.(z_1 + z_2)^{m+1} = (z_1 + z_2) \cdot (z_1 + z_2)^m.

From the inductive hypothesis, substitute the expansion for (z1+z2)m(z_1 + z_2)^m:

(z1+z2)m+1=(z1+z2)⋅∑k=0m(mk)z1m−kz2k.(z_1 + z_2)^{m+1} = (z_1 + z_2) \cdot \sum_{k=0}^m \binom{m}{k} z_1^{m-k} z_2^k.

Distribute (z1+z2)(z_1 + z_2):

(z1+z2)m+1=∑k=0m(mk)z1m+1−kz2k+∑k=0m(mk)z1m−kz2k+1.(z_1 + z_2)^{m+1} = \sum_{k=0}^m \binom{m}{k} z_1^{m+1-k} z_2^k + \sum_{k=0}^m \binom{m}{k} z_1^{m-k} z_2^{k+1}.

Step 4: Combine Terms

Adjust the indices of the second sum to match the first:

Let j=k+1j = k + 1 in the second sum. When k=0k = 0, j=1j = 1. When k=mk = m, j=m+1j = m+1. Rewrite the sum:

∑k=0m(mk)z1m−kz2k+1=∑j=1m+1(mj−1)z1m+1−jz2j.\sum_{k=0}^m \binom{m}{k} z_1^{m-k} z_2^{k+1} = \sum_{j=1}^{m+1} \binom{m}{j-1} z_1^{m+1-j} z_2^j.

Now combine the two sums:

(z1+z2)m+1=∑k=0m(mk)z1m+1−kz2k+∑j=1m+1(mj−1)z1m+1−jz2j.(z_1 + z_2)^{m+1} = \sum_{k=0}^m \binom{m}{k} z_1^{m+1-k} z_2^k + \sum_{j=1}^{m+1} \binom{m}{j-1} z_1^{m+1-j} z_2^j.

Merge the terms into a single sum:

(z1+z2)m+1=∑k=0m+1[(mk)+(mk−1)]z1m+1−kz2k.(z_1 + z_2)^{m+1} = \sum_{k=0}^{m+1} \left[\binom{m}{k} + \binom{m}{k-1}\right] z_1^{m+1-k} z_2^k.

Step 5: Pascal’s Identity

From Pascal’s Identity:

(m+1k)=(mk)+(mk−1).\binom{m+1}{k} = \binom{m}{k} + \binom{m}{k-1}.

Substitute this into the sum:

(z1+z2)m+1=∑k=0m+1(m+1k)z1m+1−kz2k.(z_1 + z_2)^{m+1} = \sum_{k=0}^{m+1} \binom{m+1}{k} z_1^{m+1-k} z_2^k.

This proves the formula for n=m+1n = m + 1.

Conclusion

By mathematical induction, the binomial formula

(z1+z2)n=∑k=0n(nk)z1n−kz2k(z_1 + z_2)^n = \sum_{k=0}^n \binom{n}{k} z_1^{n-k} z_2^k

holds for all positive integers nn.

Now you know how the Binomial Theorem works—and how mathematical induction helps us prove it! Next time you expand a binomial expression, you’ll know the math behind the magic.

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