Triangular Numbers in a ProofDate: 04/08/97 at 19:17:43 From: Suzanne Flanagan Subject: Triangular numbers in a proof 1^3 = 1^2 1^3 +2^3 = (1+2)^2 1^3 +2^3 +3^3 = (1+2+3)^2 1^3 +2^3 +3^3+ ...+n^3 = (1+2+3+...+n)^2. I am trying to prove 1^3 = 1^2 1^3 +2^3 = (1+2)^2 1^3 +2^3 +3^3 = (1+2+3)^2 1^3 +2^3 +3^3+ ...+n^3 = (1+2+3+...+n)^2. I have tried to find a proof by induction, but didn't get very far. I also tried working with triangular numbers since the right side is the triangular numbers, but I could not show that the left and right sides were equal. I need some hints, or maybe the trick that is needed to verify the general case. I also found some patterns which turned out to lead to a dead end. Please HELP! Suzanne Flanagan Date: 04/13/97 at 12:34:16 From: Doctor Steven Subject: Re: Triangular numbers in a proof The right hand side is equal to [(n*(n+1))/2]^2 so you are trying to prove that the sums of the cubes of numbers is equal to (n^4 + 2n^3 + n^2)/4. Induction should work fairly well for this proof. Start with your base case of 1: (1^4 + 2*1^3 + 1^2)/4 = 1^3 = 1. Assume it's true for k: (k^4 + 2k^3 + k^2)/4 = 1^3 + 2^3 + .... + k^3. Look at k + 1: ((k+1)^4 + 2*(k+1)^3 + (k+1)^2)/4 = (k^4 + 4k^3 + 6k^2 + 4k + 1 + 2k^3 + 6k^2 + 6k + 2 + k^2 + 2k + 1)/4= (k^4 + 6k^3 + 13k^2 + 12k + 4)/4 = (k^4 + 2k^3 + k^2)/4 + (4k^3 + 12k^2 + 12k + 4)/4 = (1^3 + 2^3 + 3^3 + . . . + k^3) + (k^3 + 3k^2 + 3k + 1) = (1^3 + 2^3 + 3^3 + . . . + k^3) + (k + 1)^3 = (1^3 + 2^3 + 3^3 + . . . + k^3 + (k+1)^3. We win! Hope this helps. -Doctor Steven, The Math Forum http://mathforum.org/dr.math/ Date: 10/09/2001 at 23:14:56 From: Madison Subject: Mathematical induction Prove by mathematical induction: 1^3+2^3+3^3+...+n^3 = (1+2+3+...+n)^2 Date: 10/10/2001 at 07:11:21 From: Doctor Floor Subject: Re: Mathematical induction Hi, Madison, It is easy to check that this is correct for n = 1. Suppose that we have proven it is correct that 1^3+2^3+3^3+...+n^3 = (1+2+3+...+n)^2, which we will use as induction hypothesis. We will use the well-known identity for triangular numbers 1+2+3+...+n = n(n+1)/2 See for instance, from the Dr. Math archives: Formula for Triangular Numbers http://mathforum.org/dr.math/problems/huff8.29.98.html Then we can derive the following: 1^3+2^3+3^3+...+n^3+(n+1)^3 = (1+2+3+...+n)^2+(n+1)^3 = (1+2+3+...+n)^2 + n(n+1)^2 +(n+1)^2 = (1+2+3+...+n)^2 + n(n+1)*(n+1) + (n+1)^2 = (1+2+3+...+n)^2 + 2(1+2+3+...+n)*(n+1) + (n+1)^2 = (1+2+3+...+n + n+1)^2 And the proof by induction is complete. If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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