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**Deadstar** Let n be a nonnegative integer. Use the identity

$\displaystyle {(1 + x)^n}{(1 + x)^n} = {(1 + x)^{2n}}$

To show that

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

So far ive gotten to...

Using the binomial theorem;

$\displaystyle (x + y)^n = \sum_{k=0}^n {n \choose k} x^{n-k}y^k$ and setting x and y to one we get;

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

so...

$\displaystyle \sum_{k=0}^n {n \choose k}^2 = (1 + x)^n(1 + x)^n = (1 + x)^{2n}$ but from here i dont know what to do... Any help please?