Proving (13^n)-(6^n) is divisible by 7 by induction

[supasupa]

Hey there evryone

I need some help with this problem as I don't 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...

 

[Office_Shredder]

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

 

[supasupa]

yeah that's what i have been trying to do but i don't know how to get

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

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

I don't know if this is the correct direction to take

 

[StatusX]

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

 

[supasupa]

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

 

[supasupa]

i get

13.13^K - 6.6^K

then what do i do??

 

[Office_Shredder]

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?

 

[StatusX]

i get

13.13^K - 6.6^K

then what do i do??

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

 

[supasupa]

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

 

[Gokul43201]

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 : $1+b+b^2+...+b^{n-1} = (1-b^n)/(1-b)$