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 Formula 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 x^{2} 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.

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