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Math Help - 2 Math Induction Questions

  1. #1
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    2 Math Induction Questions

    Ahhh...math induction...WTH is this stuff. Anyone have any good links for easy ways to learn it?

    Anyway...here are the problems:

    For ecah natural number n, with n greater than or equal to 6, 2n > (n+1)n **I was thinking since P(k) is tue, then 2k>(k+1)2 multiply both sides by two and we get: 2k*2 > 2(k2+2k+1) and I am stuck...

    2nd question:

    Prove or disprove:

    For each natural number n, 1/(1*2)+1/(2*3)+...+1/n(n+1)=n/n+1

    Any help would be great!!
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  2. #2
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    Re: 2 Math Induction Questions

    1.) Prove by induction that:

    2^n>(n+1)^2 where 6\le n\in\mathbb{N}

    First step, show that the base case P_6 is true:

    2^6>(6+1)^2

    64>49 true.

    State the induction hypothesis P_k:

    2^k>(k+1)^2

    Multiply through by 2:

    2^{k+1}>2(k+1)^2

    Now, can you show that:

    2(k+1)^2>((k+1)+1)^2 ?

    2.) Prove by induction that:

    \sum_{j=1}^n\frac{1}{j(j+1)}=\frac{n}{n+1} where n\in\mathbb{N}.

    Show that the base case P_1 is true.

    State the induction hypothesis P_k:

    \sum_{j=1}^k\frac{1}{j(j+1)}=\frac{k}{k+1}

    Now, what should you add to both sides?
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  3. #3
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    Re: 2 Math Induction Questions

    Quote Originally Posted by Measure13 View Post
    Ahhh...math induction...WTH is this stuff. Anyone have any good links for easy ways to learn it?

    Anyway...here are the problems:

    For ecah natural number n, with n greater than or equal to 6, 2n > (n+1)n **I was thinking since P(k) is tue, then 2k>(k+1)2 multiply both sides by two and we get: 2k*2 > 2(k2+2k+1) and I am stuck...

    2nd question:

    Prove or disprove:

    For each natural number n, 1/(1*2)+1/(2*3)+...+1/n(n+1)=n/n+1

    Any help would be great!!
    Think of a row of dominos. You know that any domino will push the next one over, as long as the first one is pushed.

    Mathematical induction works the same way. If you want to prove that a statement is true for all possible values, then you first need to "push the first domino" by showing the statement is true for the first value. You then need to show that "pushing any domino will push the next one over", which means that you assume the statement is true for an arbitrary value and use this to show that the statement is true for the next value.
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  4. #4
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    Re: 2 Math Induction Questions

    Quote Originally Posted by MarkFL2 View Post
    1.) Prove by induction that:

    2^n>(n+1)^2 where 6\le n\in\mathbb{N}

    First step, show that the base case P_6 is true:

    2^6>(6+1)^2

    64>49 true.

    State the induction hypothesis P_k:

    2^k>(k+1)^2

    Multiply through by 2:

    2^{k+1}>2(k+1)^2

    Now, can you show that:

    2(k+1)^2>((k+1)+1)^2 ?

    2.) Prove by induction that:

    \sum_{j=1}^n\frac{1}{j(j+1)}=\frac{n}{n+1} where n\in\mathbb{N}.

    Show that the base case P_1 is true.

    State the induction hypothesis P_k:

    \sum_{j=1}^k\frac{1}{j(j+1)}=\frac{k}{k+1}

    Now, what should you add to both sides?
    Thanks boss!
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