SolMar0306

Prove that, for all natural numbers n, 2²ⁿ + 24n − 10 is divisible by 18.

We can proceed by induction. The expression is always divisible by 2 so it is sufficient to show:
2^{2n − 1} + 12n − 5 is divisible by 9. For the base case n = 1: 2+12-5=9 so it is true. Define y = 2^{2n + 1} + 12(n + 1) − 5, which is the expression for n+1. 4(2^{2n − 1} + 12n − 5) is divisible by 9 by the induction hypothesis.

y − 4(2^{2n − 1} + 12n − 5) = 12n − 48n + 27 = − 36n + 27 = 9 * ( − 4n + 3)

y = 4(2^{2n − 1} + 12n − 5) + 9 * ( − 4n + 3)

9 | (2^{2n − 1} + 12n − 5) by the induction hypothesis and obviously divides the second term on the right so:

9 | y = 2^{2n + 1} + 12(n + 1) − 5