Sum of Consecutive CubesDate: 05/11/2000 at 20:57:35 From: cuong Subject: Sum of consecutive cube number Prove that the sum of consecutive cubes equals a square. That is: 1^3 + 2^3 + 3^3 + ... + n^3 = m^2 Here is what I did: Let n = 1: 1^3 = 1 = 1^2 (true) Let n = 2: 1^3 + 2^3 = 9 = 3^2 (true) Assume it is true for n = k: 1^3 + 2^3 + ... + k^3 = m^2 ..................[1] Prove it is true for n = k+1: 1^3 + 2^3 + ... + k^3 + (k+1)^3 = ? ..........[2] Replace [1] in [2]: m^2 + (k+1)^3 = (???) ........................[3] Consider the left-hand side of [3] for a new square. Answer by mathematical induction. Can you give me an idea? Date: 05/11/2000 at 22:14:48 From: Doctor Schwa Subject: Re: Sum of consecutive cube number The proof is a lot easier if you know the relation between m and k, so you might want to try a few examples to see if you can guess the relation, and then prove it by induction. Or, if you want me to give it away, here it is ... stop reading now if you want to figure it out for yourself ... the sum of the first k cubes is the square of the number: k(k+1)/2. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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