# Mathematical Induction

1. Jul 28, 2006

### supasupa

Hey there evryone

I need some help with this problem as I dont know which direction to go with it.

Prove by mathematical induction that (13^n)-(6^n) is divisible by 7.

The Base Step is obviously ok....

Then assume (13^K)-(6^K) is true

Then have to prove (13^(k+1))-(6^(k+1)) is true...... how do i do this

thank heaps....

2. Jul 28, 2006

### Office_Shredder

Staff Emeritus
Can you factor (13^(k+1))-(6^(k+1)) ?

3. Jul 28, 2006

### supasupa

yeah thats what i have been trying to do but i dont know how to get

(13^(k+1) - 6^(k+1))

as a multiple of (13^k - 6^k).

I dont know if this is the correct direction to take

4. Jul 28, 2006

### StatusX

$$13^{k+1}-6^{k+1}=13 \cdot (13^k-6^k)+7 \cdot 6^k$$

5. Jul 28, 2006

### supasupa

how do you do that?
like what is the step u take to get there

6. Jul 28, 2006

### supasupa

i get

13.13^K - 6.6^K

then what do i do??

7. Jul 28, 2006

### Office_Shredder

Staff Emeritus
Instead of trying that..... can you find a way to just show (13^(k+1) - 6^(k+1)) is divisible by seven? What looks like it equal seven in that equation?

8. Jul 28, 2006

### StatusX

=13.13^K-(13-7).6^K=13(13^K-6^K)+7.6^K

9. Jul 28, 2006

### supasupa

thank you very much.....that makes a heap more sense now

10. Jul 28, 2006

### Gokul43201

Staff Emeritus
In all this, you've missed out the slightly easier solution Shredder was trying to point you to. There is a standard factorization for a^n - b^n. You use a specific case of this factorization when you write the sum of a geometric series : $1+b+b^2+...+b^{n-1} = (1-b^n)/(1-b)$