1. ## Binomial coefficients

Oh someone pls help..i have hard time solving this but i can't get it...pls help me

1.show that for n greater than or equal to 0,

$\displaystyle \sum_{k=0}^{n}\binom{n}{k}^2=\binom{2n}{n}$

2. In general, the product of two polynomials with degrees m and n, respectively, is given by:

$\displaystyle \left(\sum^{m}_{i=0}a_ix^i\right)\left(\sum^{n}_{j =0}b_jx^j\right)=\sum_{r=0}^{m+n}\left(\sum_{k=0}^ {r}a_kb_{r-k}\right)x^r$

so, using $\displaystyle (x+1)^n(x+1)^m=(x+1)^{m+n}$ we have:

$\displaystyle (x+1)^{m+n}=\sum_{r=0}^{m+n}{m+n\choose r}x^r$

$\displaystyle \sum_{r=0}^{m+n}{m+n\choose r}x^r=(x+1)^m(x+1)^n=\left(\sum^{m}_{i=0}{m \choose i}x^i\right)\left(\sum^{n}_{j=0}{n \choose j}x^j\right)=$$\displaystyle \sum_{r=0}^{m+n}\left(\sum_{k=0}^{r}{m \choose k}{n \choose r-k}\right)x^r$

if
$\displaystyle \sum_{r=0}^{m+n}{m+n\choose r}x^r=\sum_{r=0}^{m+n}\left(\sum_{k=0}^{r}{m \choose k}{n \choose r-k}\right)x^r$

so:

$\displaystyle {m+n\choose r}=\sum_{k=0}^{r}{m \choose k}{n \choose r-k}$

and

$\displaystyle {n \choose k}={n \choose n-k}$

$\displaystyle {m+n\choose r}=\sum_{k=0}^{r}{m \choose k}{n \choose k}$

When m=n and r=n we have:

$\displaystyle {2n \choose n}=\sum_{k=0}^{n}{n \choose k}^2$