Proving a Binomial Sum How do I prove that $$\sum_{r=0}^{n-1}\left[ r \binom{n}{r} \binom{n}{r+1}\right]=n \binom{2n-1}{n-2}$$ without induction?
I've tried manipulating $(1+x)^n$ and the binomial coefficients, but to no avail. Also, how can one explain this identity combinatorially?

By  way of  enrichment here  is  another algebraic  proof using  basic
complex variables.

We seek to compute
$$\sum_{r=0}^n  r {n\choose r} {n\choose r+1}.$$
Introduce the integral representation
$${n \choose r+1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+2}} \; dz.$$
We  use  this to  obtain  an  integral for  the  sum.  Note that  when
$r+2>n+1$ or $r>n-1$ the pole  at zero disappears which means that the
integral is zero. Therefore we obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\sum_{r=0}^n r {n\choose r} \frac{(1+z)^n}{z^{r+2}}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^2}
\sum_{r=0}^n r {n\choose r} \frac{1}{z^r}\; dz.$$
Note that 
$$x \frac{d}{dx} (1+x)^n = \sum_{r=0}^n r {n\choose r} x^r
= nx (1+x)^{n-1}.$$
Returning to the integral this gives
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^2}
\times n \frac{1}{z} \left(1+\frac{1}{z}\right)^{n-1} \; dz
\\ = \frac{n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^3}
\frac{(1+z)^{n-1}}{z^{n-1}} \; dz
\\ = \frac{n}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{2n-1}}{z^{n+2}} \; dz.$$
This last integral can be evaluated by inspection and the result is
$$n\times {2n-1\choose n+1} = n {2n-1\choose n-2}.$$
