Math proof challenge...

[aways]

One night while bored, I "discovered" the following relation (turns out it's well known; not original to me):

The sum of the cubes of the first N integers is equal to the square of the sum of the first N integers.

For example, take N=3:

{Sum_i=1 to 3} i^3 = [{Sum_i=1 to 3}]^2

1 + 8 + 27 = [1+2+3]^2

36 = 36

Can anybody prove the general relation?

[wdfifteen]

Originally Posted by aways

One night while bored, I "discovered" the following relation (turns out it's well known; not original to me):

The sum of the cubes of the first N integers is equal to the square of the sum of the first N integers.

For example, take N=3:

{Sum_i=1 to 3} i^3 = [{Sum_i=1 to 3}]^2

1 + 8 + 27 = [1+2+3]^2

36 = 36

Can anybody prove the general relation?

Not me. This works for any value of N?

[A930Rocket]

288?

[GH85Carrera]

Originally Posted by A930Rocket

288?

No it's 2

[DanielDudley]

SO you are saying that any number is = to itself ?

[krystar]

sigma[ 1^3+2^3+....+n^3 ] = (sigma [1+...+n] ) ^ 2?

i think iteration would prob be a good approach.

1^3=1^2
1^3 + 2^3 = (1+2) ^ 2
for n+1 iteration, you're adding [n+1]^3 to left. right side would be (1+2+....+n+[n+1])^2 for the equation to work.

so in order for it to be true, (1+2+....+n+[n+1])^2 must equal to (1+2+...+n)^2 + [n+1]^3

going to try to factor out the [n+1] out of the squared expression
(1+2+....+n+[n+1])^2 =
(1+2+...+n+[n+1])*(1+2+...+n+[n+1]) =
{using (a+b)*(c+d) = ac+bc+ad+bd rule}
(1+2+...+n)*(1+2+...+n) + [n+1](1+2+...+n) + (1+2+...+n)*[n+1] + [n+1]*[n+1] =
(1+2+...+n)^2 + 2*[n+1]*(1+2+...+n) + [n+1]^2

take step back. the first term is pulled out. (sigma(n))^2. so that means the rest must equal [n+1]^3

2*[n+1]*(1+2+...+n) + [n+1]^2 ?=? [n+1]^3
{expand right}
2*[n+1]*(1+2+...+n) + [n+1]^2 ?=? [n+1]^2*[n+1]
{expand right}
2*[n+1]*(1+2+...+n) + [n+1]^2 ?=? [n+1]^2*n + [n+1]^2
{subtract [n+1]^2 term both sides}
2*[n+1]*(1+2+...+n) ?=? [n+1]^2*n
{expand right}
2*[n+1]*(1+2+...+n) ?=? [n+1]*[n+1]*n
{elimninate [n+1] factor both sides}
2*(1+2+...+n) ?=? [n+1]*n


....and i'm stuck there. who wants to finish hehe

[porsche4life]

Fuch math....

[aways]

Originally Posted by wdfifteen

This works for any value of N?

Yes.

[sammyg2]

My cat's breath smells like cat food.

[krystar]

continuing where i left off 2*(1+2+...+n) ?=? [n+1]*n

1+2+...+n = n*(n+1)/2 by Carl Gauss

2*n*(n+1)/2 ?=? [n+1]*n
{cancel out 2}
n*(n+1) ?=? [n+1]*n

YES! finished it hehe

[flatbutt]

I never understood that stuff in the general case.

[aways]

Originally Posted by krystar

continuing where i left off 2*(1+2+...+n) ?=? [n+1]*n


2*n*(n+1)/2 ?=? [n+1]*n
{cancel out 2}
n*(n+1) ?=? n*n [you mean n*(n+1) = n*(n+1)]

YES! finished it hehe

Very good. I didn't check your entire proof, but the realization that "1+2+...+n = n*(n+1)/2" by Carl Gauss is key.

[krystar]

yea saw the typo and fixed.