Official solution
Here general term = (3r + 2)20Cr
\[ \therefore \] Sum of the series = \[\sum\limits_{r = 0}^{20} {\left( {3r + 2} \right)} {}^{20}{C_r}\]
= \[3\sum\limits_{r = 0}^{20} {r.} {}^{20}{C_r} + 2\sum\limits_{r = 0}^{20} {{}^{20}{C_r}} \]
= 3 \[ \times \] 20\[ \times \]220 - 1 + 2\[ \times \]220
= 60\[ \times \]219 + 221
= 221 [15 + 1]
= 221 \[ \times \] 16
= 225