Results 1 to 3 of 3

Math Help - Proof by Induction

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    12

    Proof by Induction

    I apologize if this has been posted before. I have done many searches, but most ended up proving the Binomial Theorem. Anyhow, here goes:

    Prove 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2

    I'm basically looking for a shove in the right direction, as my induction step:

    P(k + 1) : 1^3 + 2^3 + ... + k^3 + (k + 1)^3 = ???

    I'm unsure here since the RHS can be:  (1 + 2 + ... + k + [k+1])^2 or  (2 + 3 + ... + 2k)^2

    which is where my confusion begins. Thoughts?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by flabbergastedman View Post
    I apologize if this has been posted before. I have done many searches, but most ended up proving the Binomial Theorem. Anyhow, here goes:

    Prove 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2

    I'm basically looking for a shove in the right direction, as my induction step:

    P(k + 1) : 1^3 + 2^3 + ... + k^3 + (k + 1)^3 = ???

    I'm unsure here since the RHS can be:  (1 + 2 + ... + k + [k+1])^2 or  (2 + 3 + ... + 2k)^2

    which is where my confusion begins. Thoughts?
    hi flabbergastedman,

    If 1^3+2^3+3^3+....+n^3=(1+2+3+....+n)^2

    then 1^3+2^3+3^3+....+n^3+(n+1)^3 should =[1+2+3+...+n+(n+1)]^2

    Hence

    (1+2+3+....+n+(n+1))^2 =(1+2+3+...+n)^2+(n+1)[1+2+3+...+n+(n+1)]+(1+2+3+...+n)(n+1)

    =(1+2+3+...+n)^2+2(n+1)(1+2+3+...+n)+(n+1)^2

    =(1+2+3+...+n)^2+\frac{2(n+1)(n(n+1)}{2}+(n+1)^2

    since (1+2+3+...+n)=\frac{n(n+1)}{2}

    giving

    [1+2+3+...+n+(n+1)]^2=(1+2+3+...+n)^2+n(n+1)^2+(n+1)^2

    =(1+2+3+...+n)^2+(n+1)^2(n+1)=(1+2+3+...+n)^2+(n+1  )^3

    Proven
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by flabbergastedman View Post
    I apologize if this has been posted before. I have done many searches, but most ended up proving the Binomial Theorem. Anyhow, here goes:

    Prove 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2

    I'm basically looking for a shove in the right direction, as my induction step:

    P(k + 1) : 1^3 + 2^3 + ... + k^3 + (k + 1)^3 = ???

    I'm unsure here since the RHS can be:  (1 + 2 + ... + k + [k+1])^2 or  (2 + 3 + ... + 2k)^2

    which is where my confusion begins. Thoughts?

    1^3+2^3+\ldots+k^3+(k+1)^3=\left(1+2+\ldots+k\righ  t)^2+(k+1)^3 , using the inductive hypothesis, and now use that 1+2+\ldots +n=\frac{n(n+1)}{2}

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof by induction
    Posted in the Algebra Forum
    Replies: 13
    Last Post: January 31st 2011, 05:41 PM
  2. an induction proof
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: October 5th 2009, 02:31 PM
  3. proof by induction ...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 8th 2009, 03:07 PM
  4. Mathemtical Induction Proof (Stuck on induction)
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 8th 2009, 10:33 PM
  5. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: June 8th 2008, 02:20 PM

Search Tags


/mathhelpforum @mathhelpforum