[RESOLVED] A Challenge from a Math Professor

[Code Doc]

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?

[Logophobic]

What is there to prove?

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

1³ + 2³ + 3³ + ... + n³ = (n(n+1)/2)²

[Code Doc]

[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.

[TBeck]

he's saying that if you now square both sides of

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

1³ + 2³ + 3³ + ... + n³ = (n(n+1)/2)²

[Logophobic]

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

[Code Doc]

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.