Value of Σlimits_{r = 0}^{n - 1} {frac{{^n{Cᵣ}}}{{^n{Cᵣ} + {,^n}{C

English

Gujarati

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6.Permutation and Combination

hard

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

$n + 1$

$\frac{n}{2}$

$n + 2$

None of these

Solution

(b) $\sum\limits_{r = 0}^{n – 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} = \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}}}}} $

$ = \sum\limits_{r = 0}^{n – 1} {\frac{{r + 1}}{{n + 1}}} = \frac{1}{{n + 1}}\sum\limits_{r = 0}^{n – 1} {(r + 1)} $ $ = \frac{1}{{(n + 1)}}[1 + 2 + … + n] = \frac{n}{2}$.

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