Question

Prove that 2ⁿ + 6 × 9ⁿ is always divisible by 7 for any positive integer n

Answer 1

2¹ + 6 × 9¹ = 2 + 54 = 56 is divisible by 7

When n = k,

2^k + 6 × 9^k = 7m  .....[where m is a scalar]

⇒ 6 × 9^k = 7m – 2^k …(1)

Let us prove for n = k + 1

Consider 2^k+1 + 6 × 9^k+1 = 2^k+1 + 6 × 9^k × 9

= 2^k+1 + (7m – 2^k)9 ..........(using (1))

= 2^k+1 + 63m – 9.2^k = 63m + 2^k.2¹ – 9.2^k

= 63m – 2^k (9 – 2) = 63m – 7.2^k

= 7(9m – 2^k) which is divisible by 7

∴ 2ⁿ + 6 × 9ⁿ is divisible by 7 for any positive integer n