[RESOLVED] A Challenge from a Math Professor
I was challenged last week by a math professor at a local college to prove one of "his" discoveries. He said he could not prove it, and I must admit that I am lousy at math proofs. The following statement appears true because I wrote a VB6 program for the first 100 whole numbers and it checks out:
"The sums of the cubes of the first N whole numbers equals the sums of these same N numbers squared."
e.g.,
1^3 + 2^3 = (1 + 2)^2 = 9
1^3 + 2^3 + 3^3 = (1 + 2 + 3)^2 = 36
1^3 + 2^3 + 3^3 + 4^3 = (1 + 2 + 3 + 4)^2 = 100
1^3 + 2^3 + 3^3 + 4^3 + 5^3 = (1 + 2 + 3 + 4 + 5)^2 = 225
and so on until the cows come home or even double precision overflows. Can this be proven?
Re: A Challenge from a Math Professor
What is there to prove?
1 + 2 + 3 + ... + n = n(n+1)/2
13 + 23 + 33 + ... + n3 = (n(n+1)/2)2
Re: A Challenge from a Math Professor
[n(n+1)/2]^2 = [n(n+1)/2][n(n+1)/2] = [(n^2 + n)/2][(n^2 + n)/2]
and now we have reached the 4th power of n when we evaluate this, so I guess I do not know where you are coming from.
Re: A Challenge from a Math Professor
he's saying that if you now square both sides of
Quote:
Originally Posted by Logophobic
1 + 2 + 3 + ... + n = n(n+1)/2
giving (1 + 2 + 3 + ... + n)2 = (n(n+1)/2)2
then you have the same as
Quote:
Originally Posted by Logophobic
13 + 23 + 33 + ... + n3 = (n(n+1)/2)2
Re: A Challenge from a Math Professor
You don't need to evaluate (n(n+1)/2)2
Given these summation identities:
1 + 2 + 3 + ... + n = n(n+1)/2
13 + 23 + 33 + ... + n3 = n2(n+1)2/4 = (n(n+1)/2)2
it is easy to show that
13 + 23 + 33 + ... + n3 = (1 + 2 + 3 + ... + n)2
Re: A Challenge from a Math Professor
OK, now I have it together. I recall from history that when Gauss was 11, he was acting up in his calculus class at the university. The teacher "punished" him by telling him to spend detention time after class adding up all the whole numbers from 1 to 100.
Gauss went to the blackboard and added 1 to 100 and got 101. Then he added 2 to 99 and got 101. Then he figured he would do that 50 times. So he wrote 5050 on the blackboard, underlined it, and asked to be excused.
Supposedly that was the first time anyone figured out that the sum of the first n whole numbers is n(n+1)/2.
