binomial formula for negative integer powersBinomialFormulaForNegativeIntegerPowers2005-01-20 20:10:582005-02-28 19:12:35Corollarybinomial formula% this is the default PlanetMath preamble. as your knowledge
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\addtocounter{MaxMatrixCols}{2}For negative integer powers, the binomial formula can be written in terms of binomial coefficients like so:
$$(1 - x)^{-n} = \sum_{m = 1}^\infty \binom{m+n-1}{n-1} x^m$$
{\it Proof: \,} We shall prove this by induction on $n$. First, note that, if $n=1$, then $\binom{m}{0} = 1$, so our formula reduces to
$$(1 - x)^{-1} = \sum_{m = 1}^\infty x^m ,$$
which is the formula for the sum of an infinite geometric series.
Next, suppose that the formula is valid for a certain value of $n$. Then we have
$$(1 - x)^{-n-1} = (1 - x)^{-1} (1 - x)^{-n} = \left( \sum_{k = 0}^\infty x^k \right) \left( \sum_{m = 0}^\infty {m+n-1 \choose n-1} x^m \right)$$
The product of sums can be rewritten as the following double sum:
$$\sum_{m = 0}^\infty \sum_{k = 0}^m {n+k-1 \choose n-1} x^m$$
The easiest way to see this is by rearranging the double sum as follows and adding columns
$$\begin{matrix}
x^0 \sum_{m = 0}^\infty \binom{m+n-1}{n-1} x^m = & \binom{n-1}{n-1} & + & \binom{n}{n-1} x & + & \binom{n+1}{n-1} x^2 & + & \binom{n+2}{n-1} x^3 & + & \binom{n+3}{n-1} x^4 & + & \cdots \\
x^1 \sum_{m = 0}^\infty \binom{m+n-1}{n-1} x^m = & & & \binom{n-1}{n-1} x & + & \binom{n}{n-1} x^2 & + & \binom{n+1}{n-1} x^3 & + & \binom{n+2}{n-1} x^4 & + & \cdots \\
x^2 \sum_{m = 0}^\infty \binom{m+n-1}{n-1} x^m = & & & & & \binom{n-1}{n-1} x^2 & + & \binom{n}{n-1} x^3 & + & \binom{n+1}{n-1} x^4 & + & \cdots \\
x^3 \sum_{m = 0}^\infty \binom{m+n-1}{n-1} x^m = & & & & & & & \binom{n-1}{n-1} x^3 & + & \binom{n}{n-1} x^4 & + & \cdots \\
. & . & . & . & . & . & . & . & . & . & . & .
\end{matrix}$$
To evaluate the finite sums, we shall use the following identity for binomial coefficients. (See the entry \PMlinkid{``binomial coefficient''}{273} for more information about this identity.)
$$\sum_{k = 0}^m \binom{n+k-1}{n-1} = \binom{m + n}{n}$$
Inserting this result value for the finite sum back into the double sum, we obtain
$$(1 - x)^{-n-1} = \sum_{m = 0}^\infty \binom{m + n}{n} x^m.$$
\rightline{Q.E.D.}