Prove that if [math]n \ge 0[/math], then

[[math]] \sum_{k = 0}^n {n \choose k}^2 = {{2n} \choose n}\ . [[/math]]

Hint: Write the sum as

[[math]] \sum_{k = 0}^n {n \choose k}{n \choose {n-k}} [[/math]]

and explain why this is a coefficient in the product

[[math]] (1 + x)^n (1 + x)^n\ . [[/math]]

Use this, together with Exercise, to show that

[[math]] u^{(2)}_{2n} = \frac 1{4^{2n}}{{2n}\choose n}\sum_{k = 0}^n {n \choose k}^2 = \frac 1 {4^{2n}} {{2n}\choose n}^2\ . [[/math]]