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Math Help - Proof Question: Using Mathematical Induction

  1. #1
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    Proof Question: Using Mathematical Induction

    hey guys could somebody please help me?

    i have been on this problem for a while and i dont get any where.

    The problem statement, all variables and given/known data
    Prove that, for all integers n=>1

    i get to the point where i need to prove that S(k+1) also true but what is k+1?

    is it 1/k(k+1)(k+3)


    or (k+1)(k+3)/(k+1)(k+3)(k+5)



    we had one lesson on this and this home work is damn hard :'(


    please somebody help me


    thank you ever so much
    Attached Thumbnails Attached Thumbnails Proof Question: Using Mathematical Induction-mac163.jpg  
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  2. #2
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    Feb 2008
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    Southampton, UK
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    Quote Originally Posted by boomshine57th View Post
    hey guys could somebody please help me?

    i have been on this problem for a while and i dont get any where.

    The problem statement, all variables and given/known data
    Prove that, for all integers n=>1

    i get to the point where i need to prove that S(k+1) also true but what is k+1?

    is it 1/k(k+1)(k+3)


    or (k+1)(k+3)/(k+1)(k+3)(k+5)



    we had one lesson on this and this home work is damn hard :'(


    please somebody help me


    thank you ever so much
    Firstly, I think your equation to proove is wrong. I believe it should start:
    \frac{1}{2*4}+...
    Assuming that's right, we first need to proove it is true for n=1. A quick calculation is all that's required there - we find both sides are equal to \frac{1}{8}
    Now we say "Assuming the statement is true for n=k, will it be true for n=k+1?" What we are assuming is the following:
    \sum_{r=1}^n \frac{1}{(r+1)(r+3)} = \frac{n(5n+13)}{12(n+2)(n+3)}
    What we are trying to show is that
    \sum_{r=1}^{n+1} \frac{1}{(r+1)(r+3)} = \frac{(n+1)[5(n+1)+13]}{12[(n+1)+2][(n+1)+3]}
    Now the RHS simplifies to \frac{(n+1)(5n+18)}{12(n+3)(n+4)}
    To proove, rewrite the LHS: \sum_{r=1}^n \frac{1}{(r+1)(r+3)} + \frac{1}{(n+2)(n+4)} = \frac{n(5n+13)}{12(n+2)(n+3)} + \frac{1}{(n+2)(n+4)}

    Now you just need to add the fractions, multiply out the numerator, re-factorise it and you should be there!
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  3. #3
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    Joined
    Oct 2008
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    Thumbs up

    Thank You ever so much! you are a genius

    thank you for going through all that trouble and writing it all out and thank you for noticing my mistake.

    you are a life saver
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