AL9045

What is the 1,000,000th number in this pattern?

Please include an equation for the problem.

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5...and so on.

Thanks.

Another one is what's the product for this problem?

(x+a)(x+b)(x+c)...(x+z).

Thanks.

onionmon

question -- what class is this?
i just wanan see if i can actually figure these out, cuz i dun wanna spend time if its questions for grad students

__________________
Simmons! I want you to poison Grif's next meal!
Yes sir!

I'm Dutch-Irish...
...I'm from Iowa!

AL9045

Well this is extra credit for Algebra II, but I think it's at a higher level...I hate it.

Thanks.

lordjeebus

This one's not so tricky -- just really long and time-consuming.

Here's a hint. What's (x+a)? What's (x+a)(x+b)? What's (x+a)(x+b)(x+c)? What's (x+a)(x+b)(x+c)(x+d)?

Answer those four and you should see a trend.

Edit: I should add that you'd be crazy to tackle the full 26 by hand -- do you know enough programming to write a program to do this for you?

onionmon

ALRITE, 2nd part the (x+ blah question.

use pascel's triangle
1 - 0
1 1 - 1
1 2 1 - 2
1 3 3 1 - 3
1 4 6 4 1 - 4

partial triangel. in ur case, u need to go up to level 27, but that later.

(x+a)(x+b) follows triangle level 3 --- indicated by - 2 at the side cuz ur using 2 (x+?)'s ------- x^2 + ax + bx + ab is ur answer.

as u can see, there are 1 x^2 value, 2 x values, and 1 (1) value. its more clear in the next example

(x+a)(x+b)(x+c) follows lvl 4 --- indicated by - 3 at side.

u get 1 set of x^3, 3 sets of x^2, 3 sets of x, 1 set of (1).
now u start cooking.

within the sets of x^2, all available answers are the single combinations of a,b, and c. then those multiplyed by x^2
so.. ax^2 + bx^2 + cx^2 <---- this is the (3) of the pascal triangle

second part of this is sets of x, all multiplyed combinations are with 2 letters, so u get also get 3 --- abx + bcx + cax

now. if u carry that through to the 27th level ("marked" as - 26) then u have x^27 + x^26(all single combinations of a-z, 26 of em) + x^25(all double combinations of a-z, some number) ... etc until the last number - abcdef....z

dunno if it helps... heh... i tried...

__________________
Simmons! I want you to poison Grif's next meal!
Yes sir!

I'm Dutch-Irish...
...I'm from Iowa!

onionmon

OK I GOT IT!! i was trying to figure out pattern instead of doing the problem, cuz it still helps my math in college and it helps u understand WTF is going on.

question 1.

equation ((n)(n+1))/2 is how many things there are in a series.
for example in ur case
1
2 2
3 3 3
4 4 4 4
5 5 5 5 5
etc is your series correct?
so if u do this
1 1 2 3 4 5
2 2 2 3 4 5
3 3 3 3 4 5
4 4 4 4 4 5
5 5 5 5 5 5

wat i did here was stack another triangle
1
2 2
etc
sideways on top of the first triangle
thus u get 6 units lengthwise and 5 widthwise
total 30
so because we ended at 5, we set n=5
thus 6x5 = (n+1)(n)
and because were dealing with only half of that.
((n+1)(n))/2

alrite
so u keep chucking in higher numbers cuz plugging in 5 only gives u 15 -- keep plugging in numbers for N and as soon u pass 1,000,000 -- thats the number u keep. if ur (n-1) is smaller than 1,000,000 and ur (n) is bigger than the million. that (n) is what the millionth number is.

enjoy. ask me more if u like. i gots hella time

__________________
Simmons! I want you to poison Grif's next meal!
Yes sir!

I'm Dutch-Irish...
...I'm from Iowa!

KnifeMissile

I'm sorry I didn't get to this earlier but all those thirty second mystery posts have totally ruined this forum. It didn't look like there were any intersting threads in here.

Anyway, for your first problem, onionmon is correct and the number of terms you have will be n(n+1)/2. It just so happens that n is the number in the series you are looking for! So, the equation now becomes:

n(n+1)/2 = 1,000,000

...so isolate for n and you get...

n = (sqrt(1 + 8,000,000) - 1) / 2

...or 1413.71

Now, round this up because this formula assumes we get to sum the end of the n'th line and you discover that the millionth number is 1414...


Now, the second one is also not too hard. You just have to generalize the quadratic expansion. So, let me formulate some notation 'cause I can't just scribble into the TFP...

Let SC(a-z,2) be the sum of all possible products of 2 element subsets of the variables a through z. Furthermore, define SC(a-z,0) = 1, since the null set has no elements to create a sum of products.
For example, SC(a-z,2) would look like ab + ac + ... + az + bc + bd + ... + bz + ... + yz.

Let S(i,n)f(i) denote the sum f(0) + f(1) + ... + f(i) + ... + f(n).
This will be our sigma notation.

So, the product of (x + a)(x + b)...(x + z) is equal to:

S(i,26)( SC(a-z,i) * x^(26-i) )

Pretty simple, eh?