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Series Binomial Theorem Proof for Negative Integral Powers

Page Contents

Introduction
Proof For Negative Integral n by Induction
General Conclusion

Introduction

We might think that as we have proved the Binomial Theorem for nonnegative integers, we simply put -n for n and work out the Binomials substituting -n for n. This is how we might proceed in many examples. However, the Binomial Theorem does not work the same for negative values of n as it does for nonnegative. For instance, applying the Binomial Theorem, as we might when n is positive, we get:

[1.1]

This is an infinite series, and does not converge. Clearly, we cannot always apply the binomial theorem to negative integers.

However, if the terms in a Binomial expression with negative n do converge, we can use this theorem.

In order to converge, the Binomial Theorem for numbers other than nonnegative integers, in the form (1+x)^r, requires x<1. (This was proved by Leonhart Euler).

Therefore, because the conditions for using the Binomial Theorem with powers other than nonnegative integers are different, we cannot generalise the proof for nonnegative integers to negative integers (and other real numbers).

Proof for Negative n by Induction

The proof uses the following relationship:

[2.1]

Which is the addition law with (r=-r-1). (We have proved the Addition Law for all r).

We wish to show that for n=1, ...n, and integer k≥0

[2.2]
We assume that |x|≤1

Step 1

When n=1, we have, according to our Binomial Formula:

[2.3]
By polynomial division, Method of Indeterminate Coefficients, etc, we can find:

[2.4]

We note these two equations are identical, so the Binomial Formual is true for n=1.

Step 2

We assume that if the formula is true for n=n, and that if this implies that the formula is true for n=n+1, and it is true for n=1, then it is true for all n=1, ... , n

We assume that:

[2.2, repeated]

Step 3

In order to obtain an expression for (1+x)^-(n+1), we need to divide 2.2 by (1+x), using polynomial division :

Answer:	[2.4]		Answer so far
1+x
	1+x	Subtract (1+x)	1
	...	, addition law (The justification for writing -(n+1) over zero is to follow the pattern from the other terms.) In any event, the difference above, written as a Binomial, is -(n+1) over 1. Divide (1+x) into ...
	...	The remaining x² are:
	...	The succeeding terms are each previous term times x, subtracted from the current term...
	...	In general, the terms are: Which implies the series is: which is the Binomial Expansion of -(n+1)

Because 2.4, the result of our division is what we would obtain from the Binomial Theorem, we can conclude that if the theorem is true for n, then this implies it is true for (n+1), and because it is true for n=1, then we claim to have proved the Binomial Theorem for all integers n. That is, we have proved the theorem for all negative integers, -n.

[2.2, repeated]

General Conclusion

Because we have proved the Binomial Theorem for positive n, and for negative n, we can conclude that:

[3.1]

is true for all integral n, with the provoso, then when n is negative, |x|≤1.

Ken Ward's Mathematics Pages