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Math Help - Induction Problem

  1. #1
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    Induction Problem

    This problem I could not solve

    Use induction to prove that
    3^{4n+2} + 5^{2n+1} is divisible to 14
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  2. #2
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    Quote Originally Posted by Apprentice123 View Post
    This problem I could not solve

    Use induction to prove that
    3^{4n+2} + 5^{2n+1} is divisible to 14
    Induction.
    First, we verify a "base case". (like n = 1) I'll assume n is a natural number.

    base case: (n = 1)
    3^(4+2) + 5^(2+1) = 3^7 + 5^3 = 854. (( 854/14 = 61. so we are good here)

    induction step:
    Suppose that P(n) is true. That is to say, suppose "3^(4n+2) + 5^(2n+1) is divisible by 14)" = "3^(4n+2) + 5^(2n+1) = 14k for some k" (1)
    Show P(n+1) is true.

    3^(4(n+1)+2) + 5^(2(n+1)+1)
    = 3^(4n+6) + 5^(2n+3)
    = 3^4 * 3^(4n+2) + 5^2 * 5^(2n+1)
    substitute using (1)
    = 81 * 3^(4n+2) + 25 * (14k - 3^(4n+2))
    = 81 * 3^(4n+2) - 75 *3^(4n+2) + 25*14k
    = 6 * 3^(4n+2) + 25*14k

    ...
    I've made a mistake somewhere in my algebra, b/c 6*3^(4n+2) for n=1 is not divisible by 14. sorry I can't find it (sleepy) but you get the idea?
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  3. #3
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    Quote Originally Posted by Apprentice123 View Post
    This problem I could not solve

    Use induction to prove that
    3^{4n+2} + 5^{2n+1} is divisible to 14
    I will let you check the case n=1.

    So suppose the statement is true for n = k-1, then 14| 3^{4k - 2} + 5^{2k-1}

    To prove the statement is true for n = k, check:
     3^{4k + 2} + 5^{2k+1} = 25.(3^{4k-2} + 5^{2k-1}) + 14.3^{4k-1} + 14

    From induction hypothesis, 14|25.(3^{4k-2} + 5^{2k-1}) and the rest of the terms are clearly divisible by 14. Thus 14|3^{4k + 2} + 5^{2k+1}
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  4. #4
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    My solution

    n=k
    3^{4k+2} + 5^{2k+1} = 14m
    3^{4k+6} + 5^{2k+3} = 854m
    3^{4k+6} + 5^{2k+3} = 14(61m)

    61m is integer and 3^{4k+6} + 5^{2k+3} is divisible by 14


    Are Correct ?
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  5. #5
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    Quote Originally Posted by Apprentice123 View Post
    My solution

    n=k

    3^{4k+2} + 5^{2k+1} = 14m

    3^{4k+6} + 5^{2k+3} = 854m

    You'd need to clarify that 2nd line

    3^{4k+6} + 5^{2k+3} = 14(61m)

    61m is integer and 3^{4k+6} + 5^{2k+3} is divisible by 14


    Are Correct ?
    P(k)

    3^{4k+2}+5^{2k+1}\text{\footnotesize\;\;\;is\ divisible\ by\;\;\;}14\ ?

    P(k+1)

    P(k) being true (if it is) causes

    3^{4(k+1)+2}+5^{2(k+1)+1}\text{\footnotesize\;\;\;  to\ also\ be\ divisible\ by\;\;\;}14\ ?

    Proof

    3^{4k+6}+5^{2k+3}=3^43^{4k+2}+5^25^{2k+1}

    3^4=81,\ 5^2=25, 5^2+56=3^4

    25\left(3^{4k+2}+5^{2k+1}\right)+(56)3^{4k+2}\text  {\footnotesize\;\;\;is\ divsible\ by\;\;\;}14\ ?

    56=4(14)

    therefore if P(k) is true, then it causes P(k+1) to be true.

    To find out if P(k) is true in general, test it for the initial value of n or k.
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