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# Find the next two lines of this triangle! watch

1. (Original post by zafaru123)
the outer bits are correct, but not the inner ones
Epic fail by me, I thought it was some kind of pascals triangle type thing, using nPr button on the calculator, but alas not
2. thats the general idea...however when i first solved it, i used a diff. method, which involved dividing, but it still came out correct i did for the 6th row as i only had the 5th row
6X5=30, then 30/1=30, hence 1/30
6X20=120, then because its the 2nd number=> 120/2=60, hence the 1/60
3. Or look at the denominators and divide them by row number.

1
11
121
1331
14641

I find it hard to believe that any maths teacher would be unable to solve this let alone all but one in a department.
4. It is similar to Pascal's Triangle; http://en.wikipedia.org/wiki/Pascal%27s_triangle just a slight reversal.
5. (Original post by rnd)
Or look at the denominators and divide them by row number.

1
11
121
1331
14641

I find it hard to believe that any maths teacher would be unable to solve this let alone all but one in a department.
thats what happens when you go to a school, which i think is in the bottom 400 in the country (might've been 500 not sure). but there's many methods of doing this.
6. (Original post by DeanK22)
It is similar to Pascal's Triangle; http://en.wikipedia.org/wiki/Pascal%27s_triangle just a slight reversal.
thats exactly what i thought, when i first solved it.
7. (Original post by zafaru123)
thats what happens when you go to a school, which i think is in the bottom 400 in the country (might've been 500 not sure). but there's many methods of doing this.
Schools like this offer A level maths? I'm surprised.
8. Took about 20 minutes. Next two layers are:

1/7, 1/42, 1/105, 1/140, 1/105, 1/42, 1/7
1/8, 1/56, 1/168, 1/280, 1/280, 1/168, 1/56, 1/8

Kudos to anyone else who spots the pattern (assuming I'm right).

EDIT the first: Damnit beaten to it. Congrats Tommm.
9. (Original post by rnd)
Schools like this offer A level maths? I'm surprised.
lol 5 people in a-level maths and 2 in further maths.
10. (Original post by tommm)
Spoiler:
Show

1/7 1/42 1/105 1/140 1/105 1/42 1/7

1/8 1/56 1/168 1/280 1/280 1/168 1/56 1/8

Hope there's no numerical slips, as I couldn't be bothered to go and find a calculator
I got that too.....

OP your maths dept. is retarded.
11. (Original post by zafaru123)
lol...firstly, the acid is not required...and i don't why she asked me...but only one teacher in our whole maths department got it
eugh, how do these people get into teaching maths =[
12. I've done a bit of messing around with generating functions. If we define as the bth number along in the ath row, e.g. f(1, 1) = 1, f(3, 2) = 1/6, then we find that

Furthermore, if we define , we obtain the recurrence relation

and that's where my hopes of finding a closed form expression for were dashed.
13. See my sig
14. Surely f(a+1,b+1) is just ?
15. (Original post by Elongar)
See my sig
Awesome.
16. 'A Concise Dictionary of Interesting Numbers' comes to my aid again.

ps:

1
11
21
1211

what is the limit of the ratio of the lengths of two lines?
17. (Original post by around)
'ps:

1
11
21
1211

what is the limit of the ratio of the lengths of two lines?
Spoiler:
Show
2:1. That is, a line may be no more than twice the length of the previous line.
18. , where is the unique positive root of

(Assuming no LaTeX errors, of course!)
19. Why though? I just read that somewhere and thought it was quite an amazing result. I still have no idea how Conway came up with that.

I'm guessing it comes from solving a recurrence relation based on how many characters each string generates...
20. (Original post by around)
Why though? I just read that somewhere and thought it was quite an amazing result. I still have no idea how Conway came up with that.

I'm guessing it comes from solving a recurrence relation based on how many characters each string generates...
Almost. My understanding of this (which may be incomplete) is:

There's a set of 92 "element" strings such that any string can be written as a combination of those strings, and each such string's evolution can be found by considering the evolution of the elements individually. In other words, we can write "31121112" as "3112, 1112", and the evolution of the next stage can be got by evolving "3112" and "1112" separately.

Now, what you do is make a 92 x 92 matrix M, where the (i,j) element of the M says how many elements of type "i" do you get after evolving an element of type "j".

Then if you have a 92 element column vector V saying how many you have of each element, then MV is a vector saying how many you have of each element after an iteration. (As you suggested, this is basically a recurrence relation, expressing "new V" as a linear combination of elements of "old V").

Now in the limit, we're expecting that , which you might recognize as an eigenvalue problem.

So we find an expression for the determinant of , and that turns out to be that polynomial given earlier.

Then is going to be a root of that polynomial, and it so happens it's the only positive one.

[Or something like that...]

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