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    I really get stuck with divisibility questions like this:

    Prove by induction that for N for all positive integers,

    f(n) = 2^(2n-1) + 3^(2n-1) is divisible by 5

    I'm stuck on the induction step so

    f(k+1) - f(k) = 2^(2k+1) +3^(2k+1) - 2^(2k-1) - 3^(2k-1)

    Thanks for the help
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    2^(n+2)=2^2 * 2^n=4(2^n).

    Can u use this information to rewrite some terms of your final line so u can simplify?
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    (Original post by Sajjad_K9)
    I really get stuck with divisibility questions like this:

    Prove by induction that for N for all positive integers,

    f(n) = 2^(2n-1) + 3^(2n-1) is divisible by 5

    I'm stuck on the induction step so

    f(k+1) - f(k) = 2^(2k+1) +3^(2k+1) - 2^(2k-1) - 3^(2k-1)

    Thanks for the help
    ^^^^^^.

    Sorry forgot to quote u😂
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    (Original post by Shaanv)
    ^^^^^^.

    Sorry forgot to quote u😂
    alright. so I've got this now:

    2(2^2k) + 3(3^2k) - 2^(k-1) - 3^(2k-1)

    2(2^2k) + 3(3^2k) - 2^2k(1/2) - 3^2k(1/3)

    and I'm lost again >_<
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    (Original post by Sajjad_K9)
    alright. so I've got this now:

    2(2^2k) + 3(3^2k) - 2^(k-1) - 3^(2k-1)

    2(2^2k) + 3(3^2k) - 2^2k(1/2) - 3^2k(1/3)

    and I'm lost again >_<
    Try getting everything as either 2^(2k-1) and 3^(2k-1).

    Also i think i would try f(k)+f(k+1) for this proof by induction
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    (Original post by Sajjad_K9)
    I really get stuck with divisibility questions like this:

    Prove by induction that for N for all positive integers,

    f(n) = 2^(2n-1) + 3^(2n-1) is divisible by 5

    I'm stuck on the induction step so

    f(k+1) - f(k) = 2^(2k+1) +3^(2k+1) - 2^(2k-1) - 3^(2k-1)

    Thanks for the help
    Might be clearer if you just say f(k+1)=2^{2(k+1)-1}+3^{2(k+1)-1}=2^{(2k-1)+2}+3^{(2k-1)+2}=4[2^{2k-1}]+9[3^{2k-1}]

    Then just slide the other terms into each brackets and subtract them to get:

    f(k+1)=4[2^{2k-1}+3^{2k-1}-3^{2k-1}]+9[3^{2k-1}+2^{2k-1}-2^{2k-1}]
    f(k+1)=4[f(k)-3^{2k-1}]+9[f(k)-2^{2k-1}]

    And finish it off from there using the assumption that f(k) is div. by 5
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    (Original post by RDKGames)
    Might be clearer if you just say f(k+1)=2^{2(k+1)-1}+3^{2(k+1)-1}=2^{(2k-1)+2}+3^{(2k-1)+2}=4[2^{2k-1}]+9[3^{2k-1}]
    From here I think it's simpler to calculate f(k+1) - f(k); you get

    4[2^{2k-1}]+9[3^{2k-1}] -  [2^{2k-1}]- [3^{2k-1}]  = 3[2^{2k-1}]+8[3^{2k-1}] and it's easy to split this into a mutiiple of f(k) (therefore divisible by 5) and an explicit multiple of 5.
 
 
 
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