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Is this an induction problem?

  1. Jan 30, 2012 #1
    Solved: Is this an induction problem?

    [itex](1+2+3+ \cdots + n)^2 = 1^3 + 2^3 + 3^3 + \cdots + n^3 , n \ge 1[/itex]

    Provide a derivation of the identity above.

    I do not know how to begin this problem. I tried to use induction but did not succeed. Also, I honestly do not know what it means by provide a derivation of the identity. Please do not give me the answer, I just need a helping hand in getting started.
     
    Last edited: Jan 30, 2012
  2. jcsd
  3. Jan 30, 2012 #2
    Use the fact that the sum of the first n positive integers is n(n+1)/2; then use induction.
     
  4. Jan 30, 2012 #3
    I need to show that [itex]k^3 + \big( \frac{(k+1)(k+2)}{2} \big)^2 = (k + 1)^3[/itex] or am I way off on my induction basics?

    I am following the guide I found at wolfram here http://demonstrations.wolfram.com/ProofByInduction/

    However, I cannot get the algebra to work out where f(n) + a_(n+1) = f(n+1). Thank you for the help A. Bahat.

    I keep getting a polynomial with degree four and I have no way to factor it into a cube.
     
  5. Jan 30, 2012 #4
    I would prove that 1+23+33+...+k3=k2(k+1)2/4 implies 1+23+33+...+(k+1)3=(k+1)2(k+2)2/4.
     
  6. Jan 30, 2012 #5
    I am struggling on how to manipulate [itex]k^2(k+1)^2/4 + (k+1)^3[/itex] to equal [itex](k+1)^2(k+2)^2/4[/itex]

    If I get a common denominator I get I am struggling on how to manipulate [itex](k^2(k+1)^24 + 4(k+1)^3)/4[/itex]

    However, I cannot find the route that leads to f(k+1).
     
  7. Jan 30, 2012 #6
    Use that

    [tex]\frac{k^2(k+1)^2 + 4(k+1)^3}{4}=\frac{(k+1)^2(k^2+4(k+1))}{4}[/tex]

    This is distributivity. I just did ab+ac=a(b+c).
     
  8. Jan 30, 2012 #7
    Thank you so much for your help micro and Bahat. I was overlooking distributivity. Thank you and I have it solved now. Not sure how I go about marking a thread as solved. I will try and do it by editing the title of my main post.
     
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