<s>[INST] 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?
 [/INST] 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
$$n\times {2n-1\choose n+1} = n {2n-1\choose n-2}.$$
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