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Math Help - Proof By Induction

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    Proof By Induction

    Prove that for all n +; 2^{3n+2} + 5^{n+1} is divisible by 3.


    I've learn how to prove other standard formula but don't really know how tp prove that it is divisible by an integer. Thanks in advance.
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  3. #3
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    Hello,

    Quote Originally Posted by Air View Post
    Prove that for all n +; 2^{3n+2} + 5^{n+1} is divisible by 3.


    I've learn how to prove other standard formula but don't really know how tp prove that it is divisible by an integer. Thanks in advance.
    You can do the basis

    Suppose f(n)=2^{3n+2}+5^{n+1}=3k, i.e. divisible by 3.
    \implies 2^{3n+2}=3k-5^{n+1}

    f(n+1)=2^{3(n+1)+2}+5^{(n+1)+1}=2^{3n+3+2}+5^{n+2}

    f(n+1)=8 \cdot 2^{3n+2}+5^{n+2}=8 \cdot (3k-5^{n+1})+5 \cdot 5^{n+1}

    f(n+1)=8 \cdot 3k-8 \cdot 5^{n+1}+5 \cdot 5^{n+1}

    f(n+1)=3 \cdot (8k)+5^{n+1}(-8+5)=3 \cdot (8k)-3 \cdot 5^{n+1}

    Conclude
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Air View Post
    Prove that for all n +; 2^{3n+2} + 5^{n+1} is divisible by 3.


    I've learn how to prove other standard formula but don't really know how tp prove that it is divisible by an integer. Thanks in advance.
    outline:

    Let P(n): " 2^{3n + 2} + 5^{n + 1} is divisible by 3" for all n \in \mathbb{Z}^+.

    Clearly P(1) is true (Check this!)

    Now, assume P(n) is true, that is, 2^{3n + 2} + 5^{n + 1} = 3k for some k \in \mathbb{Z} (which means P(n) is divisible by 3). we need to show that P(n + 1) is divisible by 3.

    Now, P(n + 1): 2^{3(n + 1) + 2} + 5^{(n + 1) + 1} = 8 \cdot 2^{3n + 2} + 5 \cdot 5^{n + 1}

    = 8 \cdot 2^{3n + 2} \underbrace{+ 8 \cdot 5^{n + 1} - 8 \cdot 5^{n + 1}}_{\text{this is zero}} + 5 \cdot 5^{n + 1}

    = 8(2^{3n + 2} + 5^{n + 1}) - 3 \cdot 5^{n + 1}

    ...

    i leave the last step to you


    EDIT: Hey Moo! I thought you were going to bed! you couldn't resist stealing another question from me, could you?
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