Mathematical Induction

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  • Thread starter supasupa
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  • #1
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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....
 

Answers and Replies

  • #2
Office_Shredder
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Can you factor (13^(k+1))-(6^(k+1)) ?
 
  • #3
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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
StatusX
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[tex]13^{k+1}-6^{k+1}=13 \cdot (13^k-6^k)+7 \cdot 6^k[/tex]
 
  • #5
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how do you do that?
like what is the step u take to get there
 
  • #6
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i get

13.13^K - 6.6^K

then what do i do??
 
  • #7
Office_Shredder
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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
StatusX
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supasupa said:
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
 
  • #9
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thank you very much.....that makes a heap more sense now
:smile:
 
  • #10
Gokul43201
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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 : [itex]1+b+b^2+...+b^{n-1} = (1-b^n)/(1-b) [/itex]
 

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