Official solution
\[\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}} \]
= \[{\left( {{}^{20}{C_0}} \right)^2} + {\left( {{}^{20}{C_1}} \right)^2} + {\left( {{}^{20}{C_2}} \right)^2} + .... + {\left( {{}^{20}{C_{20}}} \right)^2}\]
= 40C20
Using the formula :
\[{\left( {{}^n{C_0}} \right)^2} + {\left( {{}^n{C_1}} \right)^2} + {\left( {{}^n{C_2}} \right)^2} + .... + {\left( {{}^n{C_n}} \right)^2} = {}^{2n}{C_n}\]