Prove that 2^n + 6 x 9^n is always divisible by 7 for any positive integer n.
Answers
Answer:
2ⁿ + 6* 9ⁿ is divisible by 7
Step-by-step explanation:
Prove that 2^n + 6 x 9^n is always divisible by 7 for any positive integer n.
2ⁿ + 6* 9ⁿ
= 2ⁿ + 6 * ( 7 + 2)ⁿ
(a + b)ⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ............... + ⁿCn₋₁abⁿ⁻¹ + bⁿ
Hence
( 7 + 2)ⁿ = 7ⁿ + ⁿC₁7ⁿ⁻¹*2 + ⁿC₂7ⁿ⁻²*2² + ............... + ⁿCn₋₁7*2ⁿ⁻¹ + 2ⁿ
=2ⁿ + 6(7ⁿ + ⁿC₁7ⁿ⁻¹*2 + ⁿC₂7ⁿ⁻²*2² + ............... + ⁿCn₋₁7*2ⁿ⁻¹ + 2ⁿ)
=2ⁿ + 6*2ⁿ + (7ⁿ + ⁿC₁7ⁿ⁻¹*2 + ⁿC₂7ⁿ⁻²*2² + ............... + ⁿCn₋₁7*2ⁿ⁻¹)
= 7*2ⁿ + 7(7ⁿ⁻¹ + ⁿC₁7ⁿ⁻²*2 + ⁿC₂7ⁿ⁻³*2² + ............... + ⁿCn₋₁*2ⁿ⁻¹)
= 7 * (7ⁿ⁻¹ + ⁿC₁7ⁿ⁻²*2 + ⁿC₂7ⁿ⁻³*2² + ............... + ⁿCn₋₁*2ⁿ⁻¹ + 2ⁿ)
= 7 * (X)
where X = 7ⁿ⁻¹ + ⁿC₁7ⁿ⁻²*2 + ⁿC₂7ⁿ⁻³*2² + ............... + ⁿCn₋₁*2ⁿ⁻¹ + 2ⁿ
Hence number is divisible by 7