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Math Help - Plz help....meeeee induction

  1. #1
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    Plz help....meeeee induction

    I am studying for my 2nd midterm, but then I am stuck..plz help me..thank you so much.


    1) prove by induction that for each nonnegative integer k, the integer k^2 + 5k is even.

    2) The number n!, which is the number of permutations of the first n positive integers, is defined recursively by 0! = 1 and n! = n X (n-1)! for all n > o. Prove that n! < n ^n for all n > 0 and strict inequality holds if n > 1.
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  2. #2
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    Quote Originally Posted by jenjen View Post
    1) prove by induction that for each nonnegative integer k, the integer k^2 + 5k is even.
    I let somebody else do this problem. It is just boring.
    2) The number n!, which is the number of permutations of the first n positive integers, is defined recursively by 0! = 1 and n! = n X (n-1)! for all n > o. Prove that n! < n ^n for all n > 0 and strict inequality holds if n > 1.
    We see that,
    2!< 2^2
    Good.


    Next there is a k such as,
    k!<k^k

    But,
    k!<k^k<(k+1)^k
    Multiply by both sides by k+1>0
    Thus,
    (k+1)!<(k+1)^k(k+1)=(k+1)^{k+1}
    Q.E.D.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jenjen View Post
    1) prove by induction that for each nonnegative integer k, the integer k^2 + 5k is even.
    Let k = 0. Then 0^2 + 5 \cdot 0 = 0 is even.

    Let the theorem be true for some k = n.

    Then we need to show that the theorem is true for k = n + 1. That is:
    (n+1)^2 + 5(n+1) is even.

    So
    (n+1)^2 + 5(n+1) = n^2 + 2n + 1 + 5n + 5 = (n^2 + 5n) + (2n + 6) = (n^2 + 5n) + 2(n + 3)

    By hypothesis n^2+5n is even. The second term 2(n + 3) is a number multiplied by 2, which is even by definition. Thus the problem is in the form of an even number plus an even number, which is always even.

    -Dan
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  4. #4
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    thank you so much ThePerfectHacker and Topsquark for the quick replies!!!!!
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