MathematicsMedium64×since 2002Q2986If ∑i=120(20Ci−120Ci+20Ci−1)3=k21{\sum\limits_{i = 1}^{20} {\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}i=1∑20(20Ci+20Ci−120Ci−1)3=21k then k is equal toA100B200C50D400Check answerSkip