First slide

Combinations

Question

If $a_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}, then \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}}$ equals:

Moderate

A

$\frac{n}{2} a_{n}$
B

$\frac{n}{4} a_{n}$
C

$n a_{n}$
D

$(n - 1) a_{n}$

Solution

Let $b_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}} = \sum_{r = 0}^{n} \frac{n - r}{^{n} C_{n - r}} = \sum_{r = 0}^{n} \frac{n - r}{^{n} C_{r}}$
$\Rightarrow 2 b_{n} = \sum_{r = 0}^{n} \frac{r + (n - r)}{^{n} C_{r}} = n \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}} = n a_{n}$
$\Rightarrow b_{n} = \frac{n}{2} a_{n}$

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