Results 1 to 3 of 3

Thread: Mathematical Induction

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    54

    Exclamation Mathematical Induction

    $\displaystyle
    \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)}
    $

    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.

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

    ...


    I cant go any further!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2009
    Posts
    271
    Quote Originally Posted by Sunyata View Post
    $\displaystyle
    \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)}
    $

    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.

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

    ...


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

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

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

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

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

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

    $\displaystyle = \frac{(k + 1)(k + 4)}{4(k+2)(k+3)}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    From
    Johannesburg, South Africa
    Posts
    71
    Quote Originally Posted by Sunyata View Post
    $\displaystyle
    \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)}
    $

    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.

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

    ...


    I cant go any further!
    = $\displaystyle \frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3 )}$
    =$\displaystyle \frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}+\frac{4}{4(k+1)( k+2)(k+3)}$
    =$\displaystyle \frac{k^3+6k^2+9k+4}{4(k+1)(k+2)(k+3)}$
    =$\displaystyle \frac{(k+1)(k+1)(k+4)}{4(k+1)(k+2)(k+3)}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: Jun 29th 2010, 12:10 PM
  2. Mathematical Induction
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: Apr 7th 2010, 12:22 PM
  3. Mathematical Induction
    Posted in the Algebra Forum
    Replies: 9
    Last Post: Jul 8th 2009, 12:27 AM
  4. Mathematical Induction
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: Feb 17th 2009, 11:30 AM
  5. Mathematical Induction
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: May 30th 2007, 03:21 PM

Search Tags


/mathhelpforum @mathhelpforum