The value of sumlimits_{r = 0}^{n - 1} {frac{ | vedclass

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(B) Let $S = \sum\limits_{r = 0}^{n - 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} $.Using the property $\frac{^n{C_{r+1}}}{^n{C_r}} = \frac{n

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$\frac{n}{2}$

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(B) Let $S = \sum\limits_{r = 0}^{n - 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} $.Using the property $\frac{^n{C_{r+1}}}{^n{C_r}} = \frac{n-r}{r+1}$,we can rewrite the term inside the summation:$\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}} = \frac{1}{1 + \frac{^n{C_{r+1}}}{^n{C_r}}} = \frac{1}{1 + \frac{n-r}{r+1}} = \frac{r+1}{n+1}$.Now,substitute this back into the summation:$S = \sum\limits_{r = 0}^{n - 1} \frac{r+1}{n+1} = \frac{1}{n+1} \sum\limits_{r = 0}^{n - 1} (r+1)$.Expanding the sum: $\sum\limits_{r = 0}^{n - 1} (r+1) = 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$.Therefore,$S = \frac{1}{n+1} \cdot \frac{n(n+1)}{2} = \frac{n}{2}$.

The value of $\sum\limits_{r = 0}^{n - 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} $ equals

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