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The sum of the series $2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \ldots + 62 \cdot {}^{20}C_{20}$ is equal to:
(1) $2^{26}$ (2) $2^{25}$ (3) $2^{23}$ (4) $2^{24}$
$\sum(3r+2){}^{20}C_r = 60 \cdot 2^{19} + 2 \cdot 2^{20} = 2^{19} \cdot 64 = 2^{25}$.
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