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Section 7.2 The Generalised Binomial Theorem

We are going to present a generalised version of the special case of Theorem 3.3.2, the Binomial Theorem. In this generalisation, we will allow the exponent to be negative. Recall that the Binomial Theorem states that

\begin{equation*} (1+x)^n=\sum_{r=0}^n\binom{n}{r}x^r\text{.} \end{equation*}

If we have \(f(x)\) as in Example 7.1.2.4, we've seen that

\begin{equation*} f(x)=1/(1-x)=(1-x)^{-1}\text{.} \end{equation*}

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 \(f(x)\text{.}\)

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 7.2.1.

The generalised binomial coefficient,

\begin{equation*} \binom{n}{r}=\frac{n(n-1)\ldots(n-r+1)}{r!} \end{equation*}

where \(r \ge 0\) 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

\begin{equation*} n!/(n-r)! = n(n-1)\ldots (n-r+1) \end{equation*}

in this case.

\begin{equation*} \binom{-2}{5}=\frac{(-2)(-3)(-4)(-5)(-6)}{5!}=-6\text{.} \end{equation*}

If \(n\) is a positive integer, then we can come up with a nice formula for \(\binom{-n}{r}\text{.}\)

We have

\begin{equation*} \binom{-n}{r}=\frac{-n(-n-1)\ldots(-n-r+1)}{r!}\text{.} \end{equation*}

Taking a factor of \((-1)\) out of each term on the right-hand side gives

\begin{equation*} (-1)^rn(n+1)\ldots(n+r-1)/(r!)\text{.} \end{equation*}

Now,

\begin{equation*} (n+r-1)(n+r-2)\ldots n = (n+r-1)!/(n-1)!\text{,} \end{equation*}

so

\begin{equation*} (-1)^r\frac{n(n+1)\ldots(n+r-1)}{r!}=(-1)^r\frac{(n+r-1)!}{r!(n-1)!}=(-1)^r\binom{n+r-1}{r}\text{,} \end{equation*}

as claimed.

⬛

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

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

\begin{equation*} f(x)=(1-x)^{-1}=(1+y)^{-1}\text{,} \end{equation*}

where \(y=-x\text{.}\) The Generalised Binomial Theorem tells us that the coefficient of \(y^r\) will be

\begin{equation*} \binom{-1}{r}=(-1)^r\binom{1+r-1}{r}=(-1)^r\text{,} \end{equation*}

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

\begin{equation*} y^r=(-x)^r=(-1)^rx^r\text{,} \end{equation*}

so we have

\begin{equation*} (-1)^ry^r=(-1)^{2r}x^r=1^rx^r=x^r\text{.} \end{equation*}

Thus, the coefficient of \(x^r\) in \(f(x)\) is \(1\text{.}\) This is, indeed, precisely the sequence we started with in Example 7.1.2.4.

Let's work out \((1+x)^{-3}\text{.}\)

Solution

We need to know what \(\binom{-3}{r}\) gives, for various values of \(r\text{.}\) By Proposition 7.2.3, we have

\begin{equation*} \binom{-3}{r}=(-1)^r\binom{3+r-1}{r}=(-1)^r\binom{r+2}{r}=(-1)^r\frac{(r+2)(r+1)}{2}\text{.} \end{equation*}

When \(r=0\text{,}\) this is \((-1)^02\cdot 1/2=1\text{.}\) When \(r=1\text{,}\) this is \((-1)^13\cdot 2/2=-3\text{.}\) When \(r=2\text{,}\) this is \((-1)^24\cdot 3/2=6\text{.}\) In general, we see that

\begin{gather*} (1+x)^{-3}=1-3x+6x^2 - + \ldots +(-1)^n \frac{(n+2)(n+1)}{2}x^n+ \ldots \end{gather*}

Calculate the following.

  1. \(\displaystyle \binom{-5}{7}\)

  2. The coefficient of \(x^4\) in \((1-x)^{-2}\text{.}\)

  3. The coefficient of \(x^n\) in \((1+x)^{-4}\text{.}\)

  4. The coefficient of \(x^{k-1}\) in

    \begin{equation*} \frac{1+x}{(1-2x)^5}\text{.} \end{equation*}


    Hint

    Notice that \(\frac{1+x}{(1-2x)^5}=(1-2x)^{-5}+x(1-2x)^{-5}\text{.}\) Work out the coefficient of \(x^n\) in \((1-2x)^{-5}\) and in \(x(1-2x)^{-5}\text{,}\) substitute \(n=k-1\text{,}\) and add the two coefficients.

  5. The coefficient of \(x^k\) in \(1/(1-x^j)^n\text{,}\) where \(j\) and \(n\) are fixed positive integers.
    Hint

    Think about what conditions will make this coefficient zero.