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