LOL @ win.
to win we have to get to 200
Are the next numbers 778.5, 576 2/3, 951.5 & 692?Ok, new pattern...
86.5,
115 1/3,
259.5,
230 2/3,
432.5,
346,
605.5,
461 1/3... what are the next 2 numbers, you wont find the answer for this one on google:
Are the next numbers 778.5, 576 2/3, 951.5 & 692?
wow... how did you work that one out??
*yes, it was right
578 1512.5 3960.5 10368 27144.5ok... try this one
0.5, 2, 4.5, 12.5, 32, 84.5, 220.5...
nope this isnt right... the pattern is (x^3) - (x-1)^2
How d u get to this polynomial
? (unless you made the q but in general how do u form the pnomial)
578 1512.5 3960.5 10368 27144.5
A large amount of it was luck to be honest. It was the type of thing I could of looked at for hours and not realised, but as it happens, it took about 2 mins.wow, you are pretty good... how did you get this one? I'm sure the pattern isn't that obvious when plotted on a graph. I even took out the first number to make it look less like a Fibonacci sequence
A large amount of it was luck to be honest. It was the type of thing I could of looked at for hours and not realised, but as it happens, it took about 2 mins.
Plotting it on the chart, it looked to be in the form of of either e(x) or x^2.
The next clue was the occasion .5. Thinking that you're likely to get one from time to time if you divide a series of numbers by 2 and some happen to be odd, I multiplied everything by 2. From there it was pretty obvious they were fib^2...
ok how about this one
1 2 3 4 29
can anyone explain how to for example get the 100th number
given a series , i understand that you have to derive a polynomial
of some sort. But how to do this?
I'm fairly confident that you've not given us enough data points to define your equation...How about this
I think it's a polynomial to the power 6 or higher. Given we have only 6 data points, impossible to predict future points exactly. Can estimate though...Its a polynomial to the power of 5 or higher... i cant be bothered to work it out
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