Results 1 to 5 of 5

Math Help - Prove by induction

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    2

    Prove by induction

    Hello!
    Could somebody prove this for me?

    1+3+6+...+n(n+1)/2=n(n+1)(n+2)/6

    I mean for n=k+1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,517
    Thanks
    771
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Sylvie View Post
    Hello!
    Could somebody prove this for me?

    1+3+6+...+n(n+1)/2=n(n+1)(n+2)/6

    I mean for n=k+1

    What've you done and where're you stuck?

    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2010
    Posts
    2
    Thank you, I think I understand ;]
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Sylvie View Post
    Hello!
    Could somebody prove this for me?

    1+3+6+...+n(n+1)/2=n(n+1)(n+2)/6

    I mean for n=k+1

    P(k)

    \displaystyle\sum_{i=1}^k\frac{i(i+1)}{2}=\frac{k(  k+1)(k+2)}{6}


    P(k+1)

    \displaystyle\sum_{i=1}^{k+1}\frac{i(i+1)}{2}=\fra  c{(k+1)(k+2)(k+3)}{6}


    Proof that P(k+1) will also be true if P(k) is true

    \displaystyle\sum_{i=1}^{k+1}\frac{i(i+1)}{2}=\sum  _{i=1}^k\frac{i(i+1)}{2}+\frac{(k+1)(k+2)}{2}=\sum  _{i=1}^k\frac{i(i+1)}{2}+\frac{3(k+1)(k+2)}{6}

    which, if P(k) is valid, is

    \displaystyle\frac{k(k+1)(k+2)}{6}+\frac{3(k+1)(k+  2)}{6}=\frac{(k+1)(k+2)}{6}(k+3)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Use Induction to prove
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 11th 2012, 02:12 AM
  2. Prove the following by induction
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: November 17th 2011, 04:34 PM
  3. Replies: 10
    Last Post: June 29th 2010, 12:10 PM
  4. Prove by induction
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: February 18th 2010, 07:18 AM
  5. Prove by induction #2
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 10th 2009, 12:39 PM

Search Tags


/mathhelpforum @mathhelpforum