First slide

Binomial theorem for positive integral Index

Question

If $\sum_{r = 0}^{n} \frac{r}{^{n} C_{r}} = \sum_{r = 0}^{n} \frac{n^{2} - 3 n + 3}{2 \cdot^{n} C_{r}}$ , then

Moderate

A

n = 1
B

n = 2
C

n = 3
D

None of these

Solution

$\sum_{r = 0}^{n} \frac{r}{^{n} C_{r}} = \sum_{r = 0}^{n} \frac{n - r}{^{n} C_{n - r}} = \frac{n}{2} \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}$
$∴ \frac{n}{2} = \frac{n^{2} - 3 n + 3}{2}$
or n = 1, 3

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