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    I always get stuck on this type of question. I'd really appreciate help!

    f(n) = 2^(6n) + 3^(2n-2) is divisible by 5.

    When n = 1;
    f(1) = 65, which is divisible by 5.
    Therefore, f(n) is divisible by 5 when n=1.

    Assume that when n = k, f(k) = 2^(6k) + 3^(2k-2) is divisble by 5.

    When n = k + 1;

    f(k+1) - f(k)
    = 2^(6k+6) + 3^(2k) - 2^(6k) - 3^(2k-2)
    = 2^(6k)[64 - 1] + 3^(2k)[1 - 1/9]
    = 63.[2^(6k)] + (8/9).[3^(2k)]
    = 63.[2^(6k)] + (8).[3^(2k-2)]

    Now where do I go from here?
    Thanks! Will + rep.
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    63.[2^(6k)] + (8).[3^(2k-2)]
    = 8*[2^(6k) + 3^(2k-2)] + 55*[2^(6k)]; both terms are divisible by five, one from the assumption, the other being a multiple of 55
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    (Original post by blgeopark)
    63.[2^(6k)] + (8).[3^(2k-2)]
    = 8*[2^(6k) + 3^(2k-2)] + 55*[2^(6k)]; both terms are divisible by five, one from the assumption, the other being a multiple of 55
    You genius! Thank you so much!
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    (Original post by Narik)
    You genius! Thank you so much!
    (Original post by blgeopark)
    63.[2^(6k)] + (8).[3^(2k-2)]
    = 8*[2^(6k) + 3^(2k-2)] + 55*[2^(6k)]; both terms are divisible by five, one from the assumption, the other being a multiple of 55
    Hey Sorry to hijack the thread but I thought it might be better than having a million of these proof by induction threads:

    Similar Question:

    Show that f(k+1) - f(k) is divisible by 15.

    f(n) = 3^4n + 2^4n+2

    b) Prove that f(n) is divisible by 5

    Check for 1, assume for K etc.

    3^4(k+1) + 2^(4(k+1)+2) - (3^4k + 2^4k+2)
    = 3^(4k + 4) + 2^(4k+6) - (3^4k + 2^4k+2)
    = 81.3^4k + 16.2^(4k+2) - (3^4k + 2^4k+2)
    = 80.3^4k + 15.2^4k+2

    80 isn't divisible by 15. Stuck.

    and for part be, how do you show that it's divisible by 5? You know it's divisible by 15 which is 5 multiplied by 3 hence divisible by 5 but how do you show that?

    Sorry for hijack, thanks again!
 
 
 

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