7.2: The Generalized Binomial Theorem

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.