1. ## binomial Series Sum.

Prove that $\binom{n}{0}.\binom{n}{0}-\binom{n+1}{1}.\binom{n}{1}+\binom{n+2}{2}.\binom{ n}{2}+......................+(-1)^n.\binom{2n}{n}.\binom{n}{n}=(-1)^n$

2. Why.....

3. First we recall the identity...

$\displaystyle \sum_{i=0}^{p} \binom{m}{i}\ \binom {n}{p-i} = \binom {m+n}{p}$ (1)

... and then we consider the binomial expansion...

$\displaystyle (1+2\ x)^{n} = \sum_{k=0}^{n} \binom{n}{k} (2\ x)^{k}$ (2)

Because is...

$\displaystyle 2^{k} = \sum_{i=0}^{k} \binom{k}{i}$ (3)

... taking into account (1) we can write...

$\displaystyle (1+2\ x)^{n} = \sum_{k=0}^{n} \binom {n}{k}\ x^{k}\ \sum_{i=0}^{k} \binom {k}{i} = \sum_{k=0}^{n} \binom{n+k}{k}\ \binom{n}{k}\ x^{k}$ (4)

Now the only You have to do to obtain Your result is setting in (4) $x=-1$...

Kind regards

$\chi$ $\sigma$