sum i^2. How do you deduce the formula?

Thanks for the answers, I like both of them.

The puzzle was very interesting idea, I will check this out tomorrow.

Ben, I am sending to you my proof if you tell me how I can do this. You allow emails. But write to me and I will send the proof to you

Ana

Hey. Thanks for your proofs

2 notations remarks:

1) Instead of sum [1 to n] (i^2) I willl just write sum i^2

2) C(m,n) = m!/[n! (m-n)!] Combinatory number?

OK, this was my proof. I didnt really recall the formula, this was the reason why I had to find a way to deduce it.

I first proved (using induction) that (sum x_i)^2 = sum (x_i)^2 + 2sum [(x_i)*(x_j)], (i different from j)

Then I expressed sum i^2 this way:

1 + (1+1)^2 + (1+1+1)^2 +(1+1+1+1)^2+... +(1+1+...+1)^2 (In this last terms there are n ones)

This is equals to

1 + (1+1+2) + (1+1+1+2+2) + ... + (1+1+1+...+1+ 2+2...+2)

since any of the products (x_i)*(x_j) are 1.

In the last term, there are n ones and C(n,2) twos

1 + (2 + 2*1) + [3+ 2(1+1)] + [4 + 2 (1+1+1+...+1)] + ... + [n + 2 (1+1+1+...+1)]=

=1 + [2 + 2 C(2,2)] + [3 + 2 C(3,2)] + [4 + 2 C(4,2)] + ... + [n + 2 C(n,2)]=

= (1+2+3+4+...+n) + 2 [C(2,2)+ C(3,2)+ C(4,2)+... + C(n,2)]

The first term is n(n+1)/2

So, I just had to calculate this:

C(2,2)+ C(3,2)+ C(4,2)+... + C(n,2)

In order to use the Stieffel theorem [C(m,n) + C(m,n+1) = C(m+1,n+1)], I changed C(2,2) by C(3,3). Both combinatory numbers are equals to 1.

C(3,3) + C(3,2) + C(4,2) + C(n,5) + ... + C(n,2) =

= C(4,3) + C(4,2) + C(5,2)+ ... + C(n,2) =

= C(5,3) + C(5,2)+...+ C(n,2) =

.......

= C(n,3) + C (n,2) = C(n+1, 3)

Hence sum i^2 = n(n+1)/2 + 2 C(n+1,3)

sum i^2 = n(n+1)/2 + 2(n+1)n(n-1)/ 6

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

sum i^2 = n(n+1)(2n+1)/6

BTW, I noticed that sum i = C(n+1,2)

And sum i^2 = 2 C(n+1,3) + C (n+1,2) or

sum i^2 = C(n+1,3)+ C(n+2,3)

Funny, isnt it?

Take care and happy Valentines day to everybody!!!

Ana