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If $\sum_{i=1}^{20} \left(\frac{{}^{20}C_{i-1}}{{}^{20}C_i + {}^{20}C_{i-1}}\right)^3 = \frac{k}{21}$, then $k$ equals:
(1) 400 (2) 50 (3) 200 (4) 100
$\frac{{}^{20}C_{i-1}}{{}^{21}C_i} = \frac{i}{21}$. $\sum (i/21)^3 = \frac{(210)^2}{21^3} = \frac{100}{21}$. $k = 100$.
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