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Help needed with a likely rather obvious induction proof

  1. Dec 29, 2006 #1
    1. The problem statement, all variables and given/known data
    show that for all natural numbers n: 3^(2n+1)+2^(n-1) is divisible by 7

    3. The attempt at a solution

    i've been trying to get the second part of the proof to look like the first part, so as to be able to conclude some multiple is also divisible by 7, but i don't seem to get what needs to be done..
    3^(2(n+1)+1)+2^(n+1-1) -> 3^(2n+2+1)+2^(n+1-1) -> 3²*3^(2n+1)+2*2^(n-1) (= 7*k)
    only here i sort of get stuck trying to get the multipliers out, and i'm not certain enough of my math 'certain knowledge' otherwise to just posit that 3*(something)+2*(something) always yields multiples of 7 (not that it does, in this case)

    am i really trying to go down the wrong path here? or am i just missing something entirely too obvious? :(
  2. jcsd
  3. Dec 29, 2006 #2


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    Science Advisor

    Yes, as you say, 3*(something)+ 2*(something) does NOT always yield multiples of 7 (for example if each "something" is 1, the sum is 5 which is not a multiple of 5 so you cannot "posit" that it always does!

    What you need to show is that if 3^(2N+1)+2^(N-1) is a multiple of 7 for some specific N (do you see the difference between that and "3^(2n+1)+2^(n-1) is a multiple of 7 for all n?) then 3^(2(N+1)+1)+2^((N+1)-1) is also a multiple of 7.

    3^(2(N+1)+1)+2^(N+1-1)= 3^(2N+1+2)+2^(N+1-1)= 3²*3^(2N+1)+2*2^(N-1)= 9(3^(2N+1))+2(2^(N-1))= 2[3^(2n+1)+ 2^(N-1)]+ 7(3^(2N+1). Now, you know that 3^(2N+1)+ 2^(N-1) is a multiple of 7: 3^(2N+1)- 2^(N-1)= 7m. What does that tell you about 2[3^(2n+1)+ 2^(N-1)]+ 7(3^(2N+1)?
  4. Dec 29, 2006 #3
    ugh.. as i suspected, totally obvious :(
    thank you for the quick reply, HallsofIvy :)
    Last edited by a moderator: Dec 29, 2006
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