Show by induction that (9^n) -1 is divisibe by 8 for all nonnegative numbers.

Please help! I'm so confused with induction!

Base Case:
This is true for n = 1, because 9 - 1 = 8 is divisible by 8.

Inductive Step:
Assume that the claim is true for n = k.

Note that 9^(k+1) - 1
= 9 * 9^k - 1
= 9(9^k - 1) + 8, by arithmetic.

By inductive hypothesis, 9^k - 1 is divisible by 8.
==> 9(9^k - 1) is divisible by 8.

Certainly, 8 is divisible by 8.

Hence, their sum 9^(k+1) - 1 is also divisible by 8, completing the inductive step.

I hope this helps!

I'll have a go at this, but I don't guarantee is to be correct.

Show by induction that (9^n) -1 is divisible by 8 for all nonnegative numbers.

Note that 8|((9^a) - 1) if (9^a) - 1 = 8a for some aeZ. (*)

We proceed by induction.

BASE CASE:

Since (9^(1)) - 1 = 8, P(1) is true.

INDUCTIVE STEP:
Suppose that (9^k) - 1 is divisible by 8 such that (9^k) - 1 = 8m for some k,meZ. We will call this the inductive hypothesis. [We will show that 9^(k+1) - 1 is divisble by 8 for all keZ.]

Observe that:

9^(k+1) - 1 = 9*(9^k) - 1 = 8n (for some neZ by (*))

9^k - 1 = (8/9)n

By the inductive hypothesis, 9^k - 1 is divisible by 8, so we can express this as 9^k - 1 = 8m for some meZ.

Hence,

9^k - 1 = 8m = (8/9)n

72m = 8n

9m = n

Since 8 divides any integer multiple of 72, P(k+1) is true. Hence, (9^n) -1 is divisible by 8 for all nonnegative numbers.

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