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Series Binomial Theorem Proof for Nonnegative Integeral Powers by Induction

Series Contents

Page Contents

1. Introduction
2. Proof By Induction for Nonnegative n

Introduction

The English Mathematician, Sir Isaac Newton (1642 - 1727) presented the Binomial Theorem in all its glory without any proof. This was in the mid 1660's. The Swiss Mathematician, Jaques Bernoulli (Jakob Bernoulli) (1654 - 1705) proved the theorem by induction for nonnegative integers. Leonhart Euler (1707 - 1783), also Swiss, presented an algebraic proof for all values of n (which some claim is faulty, but this is arguable). However, Euler proved the conditions under which the Binomial Theorem converges. The Scot, Colin Maclaurin (1698 - 1746) presented a proof in calculus (which was objected to because it was calculus and not algebra). The Norwegian Mathematician, Niels Henrik Abel (1802-1829 ) finally proved the theorem for all values.

We have already proved the Binomial Theorem for nonnegative integers (not using induction), and we have proved it for all values using calculus. Here we seek to prove the theorem for nonnegative integers, this time using mathematical induction.

Proof By Induction for Nonnegative n

The essence of this proof is to use the addition formula, which we have proved for all real numbers without assuming the Binomial Theorem. We prove that the theorem is true for n=0, assume it is true for n, and show it is also true for (n+1), thus proving it by induction.

We wish to prove:
[1.1]
Where n is a nonnegative integer for values 0 to n.

Step1

We wish to prove that the theorem works for n=0.
[1.2]

Which is 1. If there is an objection to 0 over 0 being 1 (or 0 choose 0 is 1), then we can prove the theorem for n=1, 2..n. However, the Binomial Coefficient 0 over 0 is definitely 1 by definition. All the terms after term 0 are zero, because k>n

[1.3]
Similarly, all the terms after term 1 are zero because k>1.

Step2

We assume:
[1.1, repeated]

Step 3

And seek the value of (a+b)ⁿ⁺¹ by multiplying 1.1 by (a+b)
[1.2]
We multiply (a+b)ⁿ by (a+b):
[1.3]
                       [1.4]
Adding  these, term by term, we find:
[1.5]

Using the addition formula:
[1.6]

We find that if 1.1 is true then so is 1.2, completing our final induction step.












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