Question

Show that 9^(n+1) – 8n – 9 is divisible by 64, whenever n is a positive integer.

Answer 1

Solution:
n = 1 ⇒ 9^{n + 1} - 8n - 9 = 9² - 8 - 9
= 81 - 17 = 64= 1 x (64)
n = 2 ⇒9^{n + 1} - 8n - 9 = 9³ - 8(2) - 9
= 729 – 16 - 9 = 704= 11 x (64)
From n = 3, 4, 5,.....9^{n + 1} – 8n - 9 = 9^{(1 + 8)n} - 8n - 9
= 9 [nC0 + nC1 . 8 + nC2.8² + ... nCn x 8ⁿ] – 8n - 9
= 9[1 + 8n + nC2.8² + ... nCn x 8ⁿ] –8n – 9
= 9 + 72n + 9. nC2. 8² + ... 9 x nCn x 8ⁿ –8n - 9
= 8² [n + 9 (nC2 + nC3.8 +... nCn 8^n-2)]
which is divisible by 64.