Question Number 141846 by ArielVyny last updated on 24/May/21
$$\sum_{{k}\geqslant\mathrm{0}} \left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} \\ $$
Answered by mindispower last updated on 24/May/21
$$\underset{{k}\geqslant\mathrm{0}} {\sum}{C}_{{n}} ^{{k}} .{C}_{{n}} ^{{k}} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{n}} =\underset{{k}\geqslant\mathrm{0}} {\sum}\underset{{i}\geqslant\mathrm{0}} {\sum}{C}_{{n}} ^{{k}} {C}_{{n}} ^{{i}} {x}^{{k}+{i}} =\underset{{j}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}{C}_{\mathrm{2}{n}} ^{{j}} {x}^{{j}} \\ $$$${if}\:{we}\:{compar}\:{coefficient}\:{of}/{X}^{{n}} \\ $$$${C}_{\mathrm{2}{n}} ^{{n}} =\underset{{k}\geqslant\mathrm{0}} {\sum}\underset{{i}\geqslant\mathrm{0}} {\sum}{C}_{{n}} ^{{k}} {C}_{{n}} ^{{j}} ,{i}+{j}={n} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}{C}_{{n}} ^{{k}} {C}_{{n}} ^{{n}−{k}} ={C}_{\mathrm{2}{n}} ^{{n}} \Leftrightarrow\underset{{k}\geqslant\mathrm{0}} {\sum}\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} ={C}_{\mathrm{2}{n}} ^{{n}} \\ $$