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Math Help - Induction proof

  1. #1
    Senior Member I-Think's Avatar
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    Induction proof

    Good night.

    Prove by induction that 4^{2n+1}+3^{n+2} is a multiple of 13 for all values of n\geq{0}.

    All help much appreciated.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    For n=0

    4^{2n+1}+3^{n+2}=4+9=13, which is a multiple of 13.
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  3. #3
    Super Member
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    Quote Originally Posted by I-Think View Post
    Good night.

    Prove by induction that 4^{2n+1}+3^{n+2} is a multiple of 13 for all values of n\geq{0}.

    All help much appreciated.
    Base case: n = 0:
    4^{2(0)+1}+3^{0+2} = 4^1 + 3^2 = 13 OK.

    Assume that 4^{2n+1}+3^{n+2} holds true for n = k.
    4^{2k + 1} + 3^{k + 2} is divisible by 13.

    Prove that it holds true for n = k + 1:
    4^{2(k + 1) + 1} + 3^{(k + 1) +2}
    \begin{aligned}<br />
&= 4^{2k + 3} + 3^{k + 3} \\<br />
&= 16 \cdot 4^{2k + 1} + 3 \cdot 3^{k + 2} \\<br />
&= (13 + 3)\cdot 4^{2k + 1} + 3 \cdot 3^{k + 2} \\<br />
&= 13\cdot 4^{2k + 1} + 3\cdot 4^{2k + 1} + 3 \cdot 3^{k + 2} \\<br />
&= 13\cdot 4^{2k + 1} + 3(4^{2k + 1} + 3^{k + 2})<br />
\end{aligned}

    In this final sum, the first addend is obviously divisible by 13, and since what's inside the parentheses is divisible by 13, so is the second addend. Thus the final sum is divisible by 13.


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