Question (1 Viewer)

[azureus88]

Is there a way to prove this without induction?

1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6

 

[gurmies]

Yeah, there is. I think it uses cubic telescoping sums? I'm sure when Trebla gets here, he can explain it, i'm just working from memory here xD

 

[Trebla]

I've searched for the thread where I typed it in (it's quite extensive), but I think it got deleted
Basically, you consider Σ (k³ - (k - 1)³) and evaluate it in two ways. One way is by subbing the numbers in directly and you get a collapsing sum. After that, then evaluate it by expanding the cubic binomial and simplifying and separate the sums. Equate the two expressions and make Σ k² the subject.

Another quicker method (other than induction) is using something called Bernoulli polynomials and numbers...

 

[gurmies]

Taken from Math Forum - Ask Dr. Math

First note that a sum of the form

(1 - 0) + (8 - 1) + (27 - 8) + ... + (n^3 - (n-1)^3)

will collapse, since we have 1 in the first term and -1 in the second,
8 in the second and -8 in the third, and so on, to

n^3 - 0

So if we can express x^2 in terms of such differences, we can find the
sum very easily. Notice now that

x^3 - (x-1)^3 = x^3 - (x^3 - 3x^2 + 3x - 1) = 3x^2 - 3x + 1

and

x^2 - (x-1)^2 = x^2 - (x^2 - 2x + 1) = 2x - 1

We can use these to write

x^2 = ([x^3 - (x-1)^3] + 3x - 1)/3

and

x = ([x^2 - (x-1)^2] + 1)/2

We can plug the second into the first of these to get

x^2 = ([x^3 - (x-1)^3] + 3([x^2 - (x-1)^2] + 1)/2 - 1)/3

[x^3 - (x-1)^3]/3 + [x^2 - (x-1)^2]/2 - 1/6


Now, summing this for x = 1 to x = n, we get

sum(x^2) = sum([x^3 - (x-1)^3]/3 + [x^2 - (x-1)^2]/2 - 1/6)

= sum[x^3 - (x-1)^3]/3 + sum[x^2 - (x-1)^2]/2 - sum(1/6)

= [n^3 - 0^3]/3 + [n^2 - 0^2]/2 - n/6

= n^3/3 + n^2/2 - n/6

= (2n^3 + 3n^2 - n)/6