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# 7.2: The Generalized Binomial Theorem

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We are going to present a generalised version of the special case of Theorem 3.3.1, the Binomial Theorem, in which the exponent is allowed to be negative. Recall that the Binomial Theorem states that

$(1+x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r$

If we have $$f(x)$$ as in Example 7.1.2(4), we’ve seen that

$f(x) = \dfrac{1}{(1 − x)} = (1 − x)^{−1}$

So if we were allowed negative exponents in the Binomial Theorem, then a change of variable $$y = −x$$ would allow us to calculate the coefficient of $$x^n$$ in $$f(x)$$.

Of course, if $$n$$ is negative in the Binomial Theorem, we can’t figure out anything unless we have a definition for what $$\binom{n}{r}$$ means under these circumstances.

Definition: Generalised Binomial Coefficient

$\binom{n}{r} = \dfrac{n(n-1)...(n-r+1)}{r!}$

where $$r ≥ 0$$ but $$n$$ can be any real number.

Notice that this coincides with the usual definition for the binomial coefficient when $$n$$ is a positive integer, since

$\dfrac{n!}{(n − r)!} = n(n − 1). . .(n − r + 1)$

in this case.

Example $$\PageIndex{1}$$

$$\binom{−2}{5} = \dfrac{(-2)(-3)(-4)(-5)(-6)}{5!} = -6$$

If n is a positive integer, then we can come up with a nice formula for $$\binom{−n}{r}$$.

Proposition $$\PageIndex{1}$$

If $$n$$ is a positive integer, the

$\binom{−n}{r} = (-1)^r \binom{n+r-1}{r}$

Proof

We have

$$\binom{−n}{r} = \dfrac{-n(-n-1)...(-n-r+1)}{r!}$$.

Taking a factor of ($$−1$$) out of each term on the right-hand side give

$$(−1)^rn(n + 1). . . \dfrac{(n + r − 1)}{(r!)}$$.

Now,

$$(n + r − 1)(n + r − 2). . . n = \dfrac{(n + r − 1)!}{(n − 1)!}$$

so

$$(-1)^r \dfrac{n(n+1)...(n+r-1)}{r!} = (-1)^r \dfrac{(n+r-1)!}{r!(n-1)!} = (-1)^r \binom{n+r-1}{r}$$,

as claimed.

With this definition, the binomial theorem generalises just as we would wish. We won’t prove this.

Theorem $$\PageIndex{1}$$: Generalised Binomial Theorem

For any $$n ∈ R$$,

$(1+x)^n = \sum_{r=0}^{\infty} \binom{n}{r}x^r$

Example $$\PageIndex{2}$$

Let’s check that this gives us the correct values for the coefficients of $$f(x)$$ in Example 7.1.2 (4), which we already know.

Solution

We have

$$f(x) = (1 − x)^{−1} = (1 + y)^{−1}$$

where $$y = −x$$. The Generalised Binomial Theorem tells us that the coefficient of $$y^r$$ will be

$$\binom{-1}{r} = (-1)^r \binom{1+r-1}{r} = (-1)^r$$

since $$\binom{r}{r} = 1$$. But we want the coefficient of $$x^r$$, not of $$y^r$$, and

$$y^r = (−x)^r = (−1)^r x^r = x^r$$

so we have

$$(−1)^r y^r = (−1)^{2r}x^r = 1^rx^r = x^r$$

Thus, the coefficient of $$x^r$$ in $$f(x)$$ is $$1$$. This is, indeed, precisely the sequence we started with in Example 7.1.2 (4).

Example $$\PageIndex{3}$$

Let’s work out $$(1 + x)^{−3}$$.

Solution

We need to know what $$\binom{−3}{r}$$ gives, for various values of $$r$$. By Proposition 7.2.1, we have

$$\binom{-3}{r} = (-1)^r \binom{3+r-1}{r} = (-1)^r \binom{r+2}{r} = (-1)^r \dfrac{(r+2)(r+1)}{2}$$

When $$r = 0$$, this is $$(−1)^02 · \dfrac{1}{2} = 1$$. When $$r = 1$$, this is $$(−1)^13 · \dfrac{2}{2} = −3$$. When $$r = 2$$, this is $$(−1)^24 · \dfrac{3}{2} = 6$$. In general, we see that

$$(1 + x)^{−3} = 0 − 3x + 6x^2 − + . . . + (−1)^n \dfrac{(n+2)(n+1)}{2}x^n + ...$$

Exercise $$\PageIndex{1}$$

Calculate the following.

1. $$\binom{−5}{7}$$
2. The coefficient of $$x^4$$ in $$(1 − x)^{−2}$$.
3. The coefficient of $$x^n$$ in $$(1 + x)^{−4}$$.
4. The coefficient of $$x^{k−1}$$ in $\dfrac{1 + x}{(1 − 2x)^5} \nonumber$ Hint: Notice that $$\dfrac{1 + x}{(1 − 2x)^5} = (1 − 2x)^{−5} + x(1 − 2x)^{−5}$$. Work out the coefficient of $$x^n$$ in $$(1 − 2x)^{−5}$$ and in $$x(1 − 2x)^{−5}$$, substitute $$n = k − 1$$, and add the two coefficients.
5. The coefficient of $$x^k$$ in $$\dfrac{1}{(1 − x^j)^n}$$, where $$j$$ and $$n$$ are fixed positive integers. Hint: Think about what conditions will make this coefficient zero.