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}}} =

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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 have: $S = \sum\limits_{r = 0}^{n - 1} {\frac{1}{{1 + \frac{{^n{C_{r + 1}}}}{{^n{C_r}}}}}} = \sum\limits_{r = 0}^{n - 1} {\frac{1}{{1 + \frac{{n - r}}{{r + 1}}}}} $. Simplifying the denominator: $S = \sum\limits_{r = 0}^{n - 1} {\frac{{r + 1}}{{r + 1 + n - r}}} = \sum\limits_{r = 0}^{n - 1} {\frac{{r + 1}}{{n + 1}}} $. $S = \frac{1}{{n + 1}} \sum\limits_{r = 0}^{n - 1} {(r + 1)} $. Expanding the sum: $S = \frac{1}{{n + 1}} [1 + 2 + 3 + ... + n] = \frac{1}{{n + 1}} \cdot \frac{{n(n + 1)}}{2} = \frac{n}{2}$.

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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