Observe that $(1+x)^{2n}=(1+x)^n (1+x)^n.$ Hence, using binomial theorem, $\begin{align} \sum_{k=0}^{2n} {2n \choose k} x^{k} &= \left[ \sum_{k=0}^n {n \choose k} x^{k} \right]^{2} \\ &= \sum_{k=0}^{2n} \sum_{i+j=k} {k \choose i} x^i {k\choose j} x^{j} \\ &= \sum_{k=0}^{2n} \sum_{i=0}^{k} {k \choose i} {k \choose k-i} x^{k} \\ &= \sum_{k=0}^{2n} \sum_{i=0}^{k} {k \choose i}^{2} x^{k} \end{align}$