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Math Help - Mathematical Induction

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    54

    Exclamation Mathematical Induction

     <br />
\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}<br />

    Could someone please show me how to do the intermediate steps:
    Assume the results is true for n = k, where k is some positive integer.
    i.e.

    \frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)}
    R.T.P.
    \frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+4)}{4(k+2)(k+3)}
    But \frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)}
    = \frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3  )}
    =  \frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}

    ...


    I cant go any further!
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  2. #2
    Senior Member
    Joined
    Nov 2009
    Posts
    263
    Quote Originally Posted by Sunyata View Post
     <br />
\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}<br />

    Could someone please show me how to do the intermediate steps:
    Assume the results is true for n = k, where k is some positive integer.
    i.e.

    \frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)}
    R.T.P.
    \frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+4)}{4(k+2)(k+3)}
    But \frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)}
    = \frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3  )}
    =  \frac{k(k+3)^2}{4(k+1)(k+2)(k+3)} \leftarrow should be  \frac{k(k+3)^2 + 4}{4(k+1)(k+2)(k+3)}

    ...


    I cant go any further!
    = \frac{k(k+3)^2 + 4}{4(k+1)(k+2)(k+3)}

    = \frac{k^3 + 6k^2 + 9k + 4}{4(k+1)(k+2)(k+3)}

    = \frac{k^3 + 4k^2 + 2k^2 + 8k + k + 4}{4(k+1)(k+2)(k+3)}

    = \frac{k(k^2 + 2k + 1) +4(k^2 + 2k + 1)}{4(k+1)(k+2)(k+3)}

    = \frac{(k^2 + 2k + 1)(k + 4)}{4(k+1)(k+2)(k+3)}

    = \frac{(k + 1)^2(k + 4)}{4(k+1)(k+2)(k+3)}

    = \frac{(k + 1)(k + 4)}{4(k+2)(k+3)}
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  3. #3
    Junior Member
    Joined
    Sep 2009
    From
    Johannesburg, South Africa
    Posts
    71
    Quote Originally Posted by Sunyata View Post
     <br />
\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}<br />

    Could someone please show me how to do the intermediate steps:
    Assume the results is true for n = k, where k is some positive integer.
    i.e.

    \frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)}
    R.T.P.
    \frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+4)}{4(k+2)(k+3)}
    But \frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)}
    = \frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3  )}
    =  \frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}

    ...


    I cant go any further!
    = \frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3  )}
    =  \frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}+\frac{4}{4(k+1)(  k+2)(k+3)}
    =  \frac{k^3+6k^2+9k+4}{4(k+1)(k+2)(k+3)}
    =  \frac{(k+1)(k+1)(k+4)}{4(k+1)(k+2)(k+3)}
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