Let alpha = sum_{k=0}^n left( frac{({ }^n C_k | vedclass

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(D) We know that $\frac{{ }^n C_k}{k+1} = \frac{{ }^{n+1} C_{k+1}}{n+1}$.Thus,$\alpha = \sum_{k=0}^n \frac{{ }^n C_k \cdot { }^{n+1} C_{k+1}}{n+1} = \

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$10$

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(D) We know that $\frac{{ }^n C_k}{k+1} = \frac{{ }^{n+1} C_{k+1}}{n+1}$.Thus,$\alpha = \sum_{k=0}^n \frac{{ }^n C_k \cdot { }^{n+1} C_{k+1}}{n+1} = \frac{1}{n+1} \sum_{k=0}^n { }^n C_{n-k} \cdot { }^{n+1} C_{k+1} = \frac{1}{n+1} { }^{2n+1} C_{n+1}$.Similarly,$\beta = \sum_{k=0}^{n-1} \frac{{ }^n C_k \cdot { }^{n+1} C_{k+2}}{n+1} = \frac{1}{n+1} \sum_{k=0}^{n-1} { }^n C_{n-k} \cdot { }^{n+1} C_{k+2} = \frac{1}{n+1} { }^{2n+1} C_{n+2}$.Given $5 \alpha = 6 \beta$,we have $\frac{\beta}{\alpha} = \frac{5}{6}$.Substituting the expressions,$\frac{\beta}{\alpha} = \frac{{ }^{2n+1} C_{n+2}}{{ }^{2n+1} C_{n+1}} = \frac{2n+1 - (n+2) + 1}{n+2} = \frac{n}{n+2}$.Equating $\frac{n}{n+2} = \frac{5}{6}$,we get $6n = 5n + 10$,which implies $n = 10$.

Let $\alpha = \sum_{k=0}^n \left( \frac{({ }^n C_k)^2}{k+1} \right)$ and $\beta = \sum_{k=0}^{n-1} \left( \frac{{ }^n C_k \cdot { }^n C_{k+1}}{k+2} \right)$. If $5 \alpha = 6 \beta$,then $n$ equals:

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