If frac{1}{n+1} {}^{n}C_{n} + frac{1}{n} {}^{ | vedclass

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(B) The given expression is $\sum_{r=0}^{n} \frac{{}^{n}C_{r}}{r+1} = \frac{1023}{10}$.Using the identity $\frac{1}{r+1} {}^{n}C_{r} = \frac{1}{n+1} {

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$9$

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(B) The given expression is $\sum_{r=0}^{n} \frac{{}^{n}C_{r}}{r+1} = \frac{1023}{10}$.Using the identity $\frac{1}{r+1} {}^{n}C_{r} = \frac{1}{n+1} {}^{n+1}C_{r+1}$,we have:$\sum_{r=0}^{n} \frac{1}{n+1} {}^{n+1}C_{r+1} = \frac{1}{n+1} \sum_{r=0}^{n} {}^{n+1}C_{r+1}$.Let $k = r+1$,then the sum becomes $\frac{1}{n+1} \sum_{k=1}^{n+1} {}^{n+1}C_{k}$.Since $\sum_{k=0}^{n+1} {}^{n+1}C_{k} = 2^{n+1}$,we have $\sum_{k=1}^{n+1} {}^{n+1}C_{k} = 2^{n+1} - {}^{n+1}C_{0} = 2^{n+1} - 1$.Thus,$\frac{2^{n+1}-1}{n+1} = \frac{1023}{10}$.Comparing the denominators,$n+1 = 10$,which gives $n = 9$.Checking the numerator: $2^{9+1} - 1 = 2^{10} - 1 = 1024 - 1 = 1023$. This matches the given value.

If $\frac{1}{n+1} {}^{n}C_{n} + \frac{1}{n} {}^{n}C_{n-1} + \dots + \frac{1}{2} {}^{n}C_{1} + {}^{n}C_{0} = \frac{1023}{10}$,then $n$ is equal to

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