# Operation @

• Mar 17th 2010, 08:26 PM
MATNTRNG
Operation @
If operation @ is defned as: (1) 2 @ 2 = 1; (2) (2n+2) @ 2 = (2n @ 2) + (n²)/2, what is 2010 @ 2?
• Mar 18th 2010, 09:43 AM
hollywood
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.
• Mar 22nd 2010, 03:06 PM
MATNTRNG
I have been playing with this for several days and still cannot seem to put it together
• Mar 23rd 2010, 06:09 AM
hollywood
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
• Apr 1st 2010, 03:45 PM
xalk
Quote:

Originally Posted by hollywood
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

Must you not check if the above formula is correct by induction??
• Apr 2nd 2010, 12:10 AM
hollywood
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