If operation @ is defned as:(1)2 @ 2 = 1;(2)(2n+2) @ 2 = (2n@ 2) + (n²)/2, what is 2010 @ 2?

Printable View

- March 17th 2010, 08:26 PMMATNTRNGOperation @
If operation @ is defned as:

**(1)**2 @ 2 = 1;**(2)**(2*n*+2) @ 2 = (2*n*@ 2) + (*n*²)/2, what is 2010 @ 2? - March 18th 2010, 09:43 AMhollywood
If you look at the series 2 @ 2, 4 @ 2, 6 @ 2, ...

2 @ 2 = 1 = (1/2) * 2

4 @ 2 = 2 @ 2 + 1/2 = 3/2 = (1/2) * (2 + 1)

6 @ 2 = 4 @ 2 + 4/2 = 7/2 = (1/2) * (2 + 1 + 4)

8 @ 2 = 6 @ 2 + 9/2 = 8 = (1/2) * (2 + 1 + 4 + 9)

You might be interested in this formula: 1^2+2^2+3^2+.....+n^2=n*(n+1)*(2n+1)/6. I think you can get to the answer from there.

Post again in this thread if you're still having trouble. - March 22nd 2010, 03:06 PMMATNTRNG
I have been playing with this for several days and still cannot seem to put it together

- March 23rd 2010, 06:09 AMhollywood
2 @ 2 = 1 = (1/2) * 2

4 @ 2 = 2 @ 2 + 1/2 = 3/2 = (1/2) * (2 + 1)

6 @ 2 = 4 @ 2 + 4/2 = 7/2 = (1/2) * (2 + 1 + 4) = (1/2) * (2 + (1 + 4))

8 @ 2 = 6 @ 2 + 9/2 = 8 = (1/2) * (2 + 1 + 4 + 9) = (1/2) * (2 + (1 + 4 + 9))

10 @ 2 = 8 @ 2 + 16/2 = (1/2) * (2 + 1 + 4 + 9 + 16) = (1/2) * (2 + (1 + 4 + 9 + 16))

so the pattern is:

2x @ 2 = (1/2) * (2 + (1 + 4 + 9 + ... + (x-1)^2))

and you can use (with n=x-1):

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

and then set x = 1005 to get 2010 @ 2.

- Hollywood - April 1st 2010, 03:45 PMxalk
- April 2nd 2010, 12:10 AMhollywood
Yes, you should. I'm not sure whether you mean the sum-of-squares or the formula for 2x @ 2. But you can do both at once - just substitute the sum-of-squares formula into the other:

2x @ 2 = (1/2) * (2 + (x-1)*(x)*(2x-1)/6)

It should be pretty easy to do. Usually the difficult part is finding the formula - doing the proof is just working through the algebra.

- Hollywood