TAOCP 1.2.6 Exercise 49
By the binomial theorem (13), (1+x)^r=\sum_{k\ge0}\binom{r}{k}x^k, since the terms with $k<0$ vanish by definition (3).
Section 1.2.6: Binomial Coefficients
Exercise 49. [M20] Show that the identity $(1+x)^r = (1-x^2)^r(1-x)^{-r}$ implies a relation on binomial coefficients.
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Solution
By the binomial theorem (13),
$$ (1+x)^r=\sum_{k\ge0}\binom{r}{k}x^k, $$
since the terms with $k<0$ vanish by definition (3).
Similarly,
$$ (1-x^2)^r=\sum_{j\ge0}\binom{r}{j}(-1)^j x^{2j}, $$
and
$$ (1-x)^{-r}=\sum_{m\ge0}\binom{-r}{m}(-1)^m x^m. $$
Hence
$$ (1-x^2)^r(1-x)^{-r} = \left(\sum_{j\ge0}\binom{r}{j}(-1)^j x^{2j}\right) \left(\sum_{m\ge0}\binom{-r}{m}(-1)^m x^m\right). $$
Multiplying the series and collecting the coefficient of $x^n$ gives
$$ (1-x^2)^r(1-x)^{-r} = \sum_{n\ge0} \left( \sum_{j=0}^{\lfloor n/2\rfloor} (-1)^{j+n-2j} \binom{r}{j}\binom{-r}{n-2j} \right)x^n. $$
Since $(-1)^{j+n-2j}=(-1)^{n-j}$, this becomes
$$ (1-x^2)^r(1-x)^{-r} = \sum_{n\ge0} \left( \sum_{j=0}^{\lfloor n/2\rfloor} (-1)^{n-j} \binom{r}{j}\binom{-r}{n-2j} \right)x^n. $$
But the given identity states that
$$ (1+x)^r=(1-x^2)^r(1-x)^{-r}. $$
Therefore the coefficients of $x^n$ on both sides are equal. Hence
$$ \boxed{ \binom{r}{n} = \sum_{j=0}^{\lfloor n/2\rfloor} (-1)^{,n-j} \binom{r}{j}\binom{-r}{n-2j} }. $$
This relation holds for all real $r$ and all integers $n\ge0$.
This completes the proof.
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