Solve This Pattern

[DrPizza]

When speaking about quantum mechanics, our physics prof used to do this with us:

Solve for this series:

2, 4, 6, 8, 10, ...x

We would all say "12!"

He would say WRONG!, it's "A"

We would say "WTF???"

And he explained:

The entire series is 2, 4, 6, 8, 10, A, 14, 16, 18, 20, 22, B....

You guessed wrong because you didn't look beyond the unknown to get a clue as to what you are looking at. Particle physics is a lot like this. You can not assume anything, because at the quantum level, a whole new series of rules apply than those we are used to. Due to a reason that is 6 steps down the road, there may be a completely unexpected variance that will change your assumed result.

This has blown my mind ever since.

I think my favorite pattern comes from putting points on the circumference of a circle, connecting every combination of two points with a straight line segment, and count how many areas the circle is divided into.
With 1 point, there is no segment, so clearly the circle has only one region.
With 2 points, there is 1 segment, and the circle is divided into two regions.
3 points, 3 segments, 4 regions.

Continuing,
1 point 1 region
2 points 2 regions
3 points 4 regions
4 points 8 regions
5 points 16 regions

SOooooooo, if we have 6 points, how many regions should there be? Okay, draw it and check it. If it doesn't work at first, you might need to space out your points a little differently along the circumference. "Don't spend more than 10 or 15 minutes working on this tonight; some people just seem to never be able to do it, no matter how much time they spend."

I can be such a jerk sometimes... it's impossible to get 32 regions that way. The number of regions a circle is divided into is 1 + the number of segments + the number of intersections of those segments. Each segment increases the regions by 1, each time a segment crosses another segment, the number of regions increases by 1. Since it takes 2 points to make a segment, and 4 points to have 2 segments which intersect inside the circle, the number of regions, x, is given by
x=1+ nC2 + nC4 where C is the combination of two elements taken from n points. The sequence goes like this: 1,2,4,8,16,31
Ooops! 31?! Bahahaha! Sorry kids!

 

[CPA]

http://oeis.org/search?q=20,+125,+200&language=english&go=Search

The online encyclopedia of integer sequences. Apparently, your sequence is uninteresting.

So, alternative #2: since there are 4 points, simply run a cubic regression for a best fit curve. With 4 points, the curve can be made to fit those 4 values perfectly.

x=3.75n^3 - 37.5n² + 191.25n - 137.5
for n=1, x=20
for n=2, x=125
for n=3, x=200
and for n=5, x=350, as required by your sequence.

For n=4, x=267.5
and for n=6, x=470

giving you 20,125,200,267.5,350,470
-----

However, you can make up any value you want for the 4th term, and use a quartic regression.

Thus, you can have 20,125,200,700,350, and after running a quartic regression, and finding the value for that quartic regression for the 6th term, you wind up with -3855

So, a perfectly valid answer is 20,125,200,700,350,-3855
This corresponds to the equation x = -72.08333333333n^4+796.666666666n^3-2992.91666666666n^2+4588.3333333n-2300

This equation has the nice feature that x's will always (it appears; I can prove whether this is true or not, but that would cost you) be integers. Those are repeating decimals in the equation; lacking formatting, I chose to write them that way (rather than -2992 11/12 n^2, which isn't as clear.)

The next terms are -16650, -44500, -95600, -179875, -308980, -496300, -756950

What do I win?

Nerd

 

[KMc]

20, 125, 200, 290, 350, 425

This.

 

[Cuda1447]

This thread is to smart for me....


*slowly backs away with hands in the air*

 

[JDawg1536]

is there an answer or are you trolling?

It's just a problem my friend posted. It is supposedly a 5th grade level problem, but it had me stumped. I don't know what the answer is supposed to be.

 

[DrPizza]

Ahhhh, a 5th grade problem. That means the teacher likely doesn't know what the answer is supposed to be other than it's supposed to match the answer the key in the back of the teacher's edition. As many of us have pointed out, there are an endless number of *correct* answers to this problem.

What is more important is that given such an open question, a student is able to justify their answer. Unfortunately, that's not the way math is generally taught. (Though, that's about to change with the implementation of the new common core in 40-something of the states.)

 

[JDawg1536]

Ahhhh, a 5th grade problem. That means the teacher likely doesn't know what the answer is supposed to be other than it's supposed to match the answer the key in the back of the teacher's edition. As many of us have pointed out, there are an endless number of *correct* answers to this problem.

What is more important is that given such an open question, a student is able to justify their answer. Unfortunately, that's not the way math is generally taught. (Though, that's about to change with the implementation of the new common core in 40-something of the states.)

Thanks for your explanations... those were a bit over my head, but very interesting.

 

[Noirish]

20 125 200 335 350 605

 

[shortylickens]

Fucking calculus. Fucking derivitives. Fucking limits.

FUCK EM!!!!