If ∑k=131 31Ck 31Ck−1−∑k=130 30Ck 30Ck−1=α(60!)(30!)(31!), where α∈ℝ, then the value of 16α is equal to

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If $\sum_{k = 1}^{31} (^{31} C_{k}) (^{31} C_{k - 1}) - \sum_{k = 1}^{30} (^{30} C_{k}) (^{30} C_{k - 1}) = \frac{α (60!)}{(30!) (31!)}$ , where $α \in ℝ$ , then the value of $16 α$ is equal to

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1320

1615

1411

1855

answer is A.

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Detailed Solution

$\begin{array}{l} \sum_{k = 1}^{31}^{31} C_{k} \cdot^{31} C_{k - 1} =^{31} C_{1} \cdot^{31} C_{0} +^{31} C_{2} \cdot^{31} C_{1} + \dots . +^{31} C_{31} \cdot^{31} C_{30} \\ =^{31} C_{0} \cdot^{31} C_{30} +^{31} C_{1} \cdot^{31} C_{29} + \dots +^{31} C_{30} \cdot^{31} C_{0} =^{62} C_{30} \end{array}$

$\begin{array}{l} \sum_{k = 1}^{30} (^{30} C_{k} \cdot^{30} C_{k - 1}) =^{60} C_{29} \\ \sum_{k = 1}^{31} (^{31} C_{k}) (^{31} C_{k - 1}) - \sum_{k = 1}^{30} (^{30} C_{k}) (^{30} C_{k - 1}) =^{62} C_{30} -^{60} C_{29} = \frac{62!}{30! 32!} - \frac{60!}{29! 31!} \\ \Rightarrow \frac{α (60!)}{(30!) (31!)} = \frac{60!}{29! 31!} \{\frac{62 \cdot 61}{30 \cdot 32} - 1\} \\ \Rightarrow \frac{α (60!)}{(30!) (31!)} = \frac{60!}{30! 31!} (\frac{2822}{32}) \\ ∴ 16 α = 16 \times \frac{2822}{32} = 1411 \end{array}$