First slide

Binomial theorem for positive integral Index

Question

If $\sum_{r = 0}^{n} (pr + 2) \cdot^{n} C_{r} = (25) (64)$ where n, $p \in N$ , then

Moderate

A

p = 3
B

p = 4
C

n = 7
D

n = 6

Solution

$S = \sum_{r = 0}^{n} (pr + 2) \cdot^{n} C_{r}$
$∴ = \sum_{r = 0}^{n} (p (n - r) + 2) \cdot^{n} C_{r}$ $(∵^{n} C_{r} =^{n} C_{n - r})$
Adding these, we get
$2 S = (pn + 4) \sum_{r = 0}^{n}^{n} C_{r}$
$\Rightarrow S = (pn + 4) \times 2^{n - 1}$
So, n = 7 and 7p + 4 = 25 or p = 3.

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