If { }^{n} C_0+frac{1}{2}{ }^{n} C_1+frac{1}{ | vedclass

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(C) We know the identity $\frac{1}{k+1}{ }^{n} C_k = \frac{1}{n+1}{ }^{n+1} C_{k+1}$.Substituting this into the given sum:$\sum_{k=0}^{n} \frac{1}{k+1

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$9$

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(C) We know the identity $\frac{1}{k+1}{ }^{n} C_k = \frac{1}{n+1}{ }^{n+1} C_{k+1}$.Substituting this into the given sum:$\sum_{k=0}^{n} \frac{1}{k+1}{ }^{n} C_k = \sum_{k=0}^{n} \frac{1}{n+1}{ }^{n+1} C_{k+1} = \frac{1}{n+1} \sum_{k=0}^{n} { }^{n+1} C_{k+1}$.Let $j = k+1$,then the sum becomes $\frac{1}{n+1} \sum_{j=1}^{n+1} { }^{n+1} C_j$.Since $\sum_{j=0}^{n+1} { }^{n+1} C_j = 2^{n+1}$,we have $\sum_{j=1}^{n+1} { }^{n+1} C_j = 2^{n+1} - { }^{n+1} C_0 = 2^{n+1} - 1$.Thus,$\frac{2^{n+1}-1}{n+1} = \frac{1023}{10}$.Comparing the denominators,$n+1 = 10 \implies n = 9$.Checking the numerator: $2^{9+1} - 1 = 2^{10} - 1 = 1024 - 1 = 1023$.Therefore,$n = 9$.

If ${ }^{n} C_0+\frac{1}{2}{ }^{n} C_1+\frac{1}{3}{ }^{n} C_2+\ldots+\frac{1}{n+1}{ }^{n} C_{n}=\frac{1023}{10}$,then $n=$

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