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.