MathematicsHard64×since 2002Q3001If ∑k=131(31Ck)(31Ck−1)−∑k=130(30Ck)(30Ck−1)=α(60!)(30!)(31!)\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} }k=1∑31(31Ck)(31Ck−1)−k=1∑30(30Ck)(30Ck−1)=(30!)(31!)α(60!), where α\alphaα ∈\in∈ R, then the value of 16α\alphaα is equal toA1411B1320C1615D1855Check answerSkip