If sumlimits_{k=1}^{31} binom{31}{k} binom{31 | vedclass

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(D) We know that $\sum\limits_{k=1}^{n} \binom{n}{k} \binom{n}{k-1} = \binom{2n}{n-1}$.Applying this to the first term with $n=31$,we get $\sum\limits

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$1855$

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(D) We know that $\sum\limits_{k=1}^{n} \binom{n}{k} \binom{n}{k-1} = \binom{2n}{n-1}$.Applying this to the first term with $n=31$,we get $\sum\limits_{k=1}^{31} \binom{31}{k} \binom{31}{k-1} = \binom{62}{30}$.Applying this to the second term with $n=30$,we get $\sum\limits_{k=1}^{30} \binom{30}{k} \binom{30}{k-1} = \binom{60}{29}$.Thus,the expression becomes $\binom{62}{30} - \binom{60}{29} = \frac{62!}{30!32!} - \frac{60!}{29!31!}$.Factor out $\frac{60!}{29!31!}$: $\frac{60!}{29!31!} \left( \frac{62 	imes 61}{30 	imes 32} - 1 ight) = \frac{60!}{30!31!} \left( \frac{3782}{32} ight) = \frac{60!}{30!31!} \left( \frac{1891}{16} ight)$.Comparing this with $\frac{\alpha(60!)}{(30!)(31!)}$,we get $\alpha = \frac{1891}{16}$.Therefore,$16\alpha = 1891$.

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

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