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### Proof by Induction

```
Date: 06/15/99 at 16:40:02
From: Sara Bergman
Subject: Show by induction

Show by induction that  (4^n)-1  is divisible by 3 for all n >= 1.

So far so good...?
n = 1   Left = Right

n = p+1
(4^p+1)-1 = 3k (k = constant)
4*(4^p)-1 = 3k
After that...?
```

```
Date: 06/15/99 at 17:55:38
From: Doctor Anthony
Subject: Re: Show by induction

Suppose  f(n) =  4^n - 1

Let M(3) stand for 'is a multiple of 3'

We must prove that f(n) = M(3) for all n >= 1.

Check that it is true for n = 1, i.e. 4^1 - 1 = 3 = M(3). So it's true
for n = 1.

Assume it's true for n = k. That is, we assume  4^k - 1 = M(3).

Now consider n = k+1

f(k+1) = 4^(k+1) - 1

= 4*4^k + (-4 + 4) - 1

= (4*4^k - 4) + 4 - 1

= 4(4^k - 1) + 4 - 1

= 4(4^k - 1) + 3

= M(3) + M(3)

= M(3)

So if f(k) is M(3), then so is f(k+1). Since f(1) is M(3), so then
f(2) is also M(3). And if f(2) is M(3), then f(3) is M(3). And if f(3)
is M(3), then f(4) is M(3), and so on to all positive integer values
of n.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Discrete Mathematics

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