Question

If n is a positive integer, using Binomial theorem, show that, 9ⁿ⁺¹ − 8n − 9 is always divisible by 64

Answer 1

(1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ........ + ⁿC_n−1 xⁿ⁻¹ + ⁿC_nxⁿ 

Put x = 8 we get

(1 + 8)ⁿ = ⁿC₀ + ⁿC₁(8) + ⁿC₂(8)² + ......... + ⁿC_n–1 8^n–1 + ⁿC_n . 8ⁿ 

9n = ⁿC₀ + ⁿC₁(8) + ⁿC₂(8)² + ......... + ⁿC_n–1 8^n–1 + ⁿC_n . 9ⁿ = 1 + 8n + ⁿC₂ × 8² + ........ +

ⁿC_n–1 8^n–1 + ⁿC_n . 8ⁿ

9ⁿ - 8n – 1 = ⁿC₂ × 8² + ........ + ⁿC_n–1 8^n–1 + ⁿC_n × 8ⁿ

9ⁿ - 8n – 1 = 8² [ⁿC₂ + .......... + ⁿC_n–1 × 8^n–3 + ⁿC_n × 8^n–2]

Which is divisible by 64 for all positive integer n.

∴ 9ⁿ – 8n – 1 is divisible by 64 for all positive integer n.

Put n = n + 1 we get

9^{n + 1} – 8 (n + 1) – 1 is divisible by 64 for all possible integer n

(9^{n + 1} – 8n – 8 – 1) is divisible by 64

∴ 9^{n + 1} – 8n – 9 is always divisible by 64