Official solution
\[\sum\limits_{r = 0}^{20} {{r^2}.{}^{20}{C_r}} \]
\[\sum {(4(r - 1) + r).{}^{20}{C_r}} \]
\[\sum {r(r - 1).{{20 \times 19} \over {r(r - 1)}}} .{}^{18}{C_r} + r.{{20} \over r}.\sum {{}^{19}{C_{r - 1}}} \]
\[ \Rightarrow 20 \times {19.2^{18}} + {20.2^{19}}\]
\[ \Rightarrow 420 \times {2^{18}}\]