First slide

Binomial theorem for positive integral Index

Question

For all n€ R, $9^{n + 1}$ - 8n - 9 is divisible by

Moderate

A

64
B

63
C

62
D

65

Solution

$\begin{aligned} 9^{n + 1} - 8 n - 9 = 9^{n} \times 9 - 8 n - 9 \\ = (1 + 8)^{n} \times 9 - 8 n - 9 \\ = (^{n} C_{0} +^{n} C_{1} 8 +^{n} C_{2} 8^{2} +^{n} C_{3} 8^{3} + \dots +^{n} C_{n} 8^{n}) 9 - 8 n - 9 \end{aligned}$
$\begin{aligned} = (1 + 8 n +^{n} C_{2} 8^{2} +^{n} C_{3} 8^{3} + \dots +^{n} C_{n} 8^{n}) 9 - 8 n - 9 \\ = 9 + 72 n + (^{n} C_{2} 8^{2} +^{n} C_{3} 8^{3} + \dots +^{n} C_{n} 8^{n}) 9 - 8 n - 9 \\ = (72 n - 8 n) + 8^{2} (^{n} C_{2} +^{n} C_{3} 8 + \dots +^{n} C_{n} 8^{n - 2}) 9 \\ = 64 n + 64 (^{n} C_{2} +^{n} C_{3} 8 + \dots +^{n} C_{n} 8^{n - 2}) 9 \\ = 64 [n + (^{n} C_{2} +^{n} C_{3} 8 + \dots +^{n} C_{n} 8^{n - 2}) 9] \end{aligned}$
= 64 x Some constant numbers
= Divisible by 64

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