We have two responses fo you

Hi James.

This is an induction proof. We have several examples in our archive.

The basic method of induction proofs is this:

1. Prove the hypothesis is true for certain small value(s) of n.
2. Demonstrate that if the hypothesis is true for n, it is also true for n+1.

So to do part 1, you simply show that for n = 1, the value is indeed divisible by 11. That's just arithmetic.

For part 2, you use (n+1) in place of n in the expression. Then you break it up in such a way that you can get the expression that had just n in it as a component. I'll show you what I mean:

(27)23⁽ⁿ⁺¹⁾ + (17)10²⁽ⁿ⁺¹⁾

What I want is to transform this into an expression that has (27)23ⁿ + (17)10²ⁿ in it.

(27)23⁽ⁿ⁺¹⁾ + (17)10²⁽ⁿ⁺¹⁾
=(27)23ⁿ(23¹) + (17)10²ⁿ(10²)
=(27)23ⁿ(22+1) + (17)10²ⁿ(99+1)
= (27)23ⁿ + (17)10²ⁿ + (27)23ⁿ(22) + (17)10²ⁿ(99)

Now you can see the original (27)23ⁿ + (17)10²ⁿ in there (which we know is divisible by 11) and you just have to show that the remaining terms are also divisible by 11.

Cheers,
Stephen La Rocque.

James,

Use arithmetic modulo 11.

Penny