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How do I model this problem with an equation?...
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File: Capture.png (78 KB, 697x232) Image search: [iqdb] [SauceNao] [Google]
78 KB, 697x232
How do I model this problem with an equation? I made a table of values and I get the pattern, but not sure what the equation is.
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https://en.wikipedia.org/wiki/Triangular_number
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How would I go about figuring that equation out though? Is there a certain way to do so, other than trial and error?
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>>51062
x = 1 + n*1
where n is the index number.

gg ez wp
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>>51062
Well, you write down the first numbers (note the different colors of the balls):
1
1+2 = 3
1+2+3 = 6
1+2+3+4 = 10
So the 50th triangular number is the sum of the first 50 natural numbers. If you don't know that this is equal to n choose 2 = n(n+1)/2 for n=50 you can prove it as follows just like Gauss:

Write the numbers that you want to sum in ascending order and below in descending order:
1 2 3 ... n-2 n-1 n
n n-1 n-2 ... 3 2 1

The sum in a column is always n+1 so the sum of all 2n numbers is n(n+1). Since you only want to count each number once, you need to divide by 2:

T_n = n choose 2 = n(n+1)/2
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>>51081
You can also do this with the balls if you have problems with the equation:

Bring the Tn balls in a right triangle pattern:
*
**
***
****

Use another triangle with the same number of balls, but in reverse order, and put the two of them together:

* ****
** ***
*** **
**** *
You obtain a rectangle with n rows and n+1 balls in each row, that is, n(n+1) balls in total. Since you used two triangles, the number of balls in Tn is half this amount, n(n+1)/2.
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File: Capture.png (33 KB, 683x236) Image search: [iqdb] [SauceNao] [Google]
33 KB, 683x236
>>51081
>>51092

Thanks, that helped.

How would I apply the ascending/descending thing to a problem like this?
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>>51095
It's only the application of the triangular numbers before. You have numbered balls / squares, so in the first rows:
1
23
456
and so on. Note that the rightmost number is equal to T_n for n=1,2,3...

Since in the problem you have 60 completely filled rows you have used the balls numbered from 1 to T(60).

In the 61st row are the balls numbered T(60)+1,..., T(60)+23 (which you need), and so on.
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>>51100
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File: Capture.png (151 KB, 1292x233) Image search: [iqdb] [SauceNao] [Google]
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>>51100
Ahh, didn't realize it was the same as the other problem.

And one final question about pic related. I found the equation for this problem to be:

t(n)=1/2^(n-1)

How do I look at this in terms of the general equation? What is my "a" and "r" ?
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>>51115
>t(n)=1/2^(n-1)
You can write this as (1/2)^(n-1) so a = 1 (the starting number t(1) = 1 = a ) and r = 1/2.

For the sum (part iii) ) you need to use an approach similar to the one with the triangular numbers.

Write the terms that you want to add, say up to t(n)

a + ar + ar^2 + ar^3 + .... + ar^(n-1) = s = s(n) (1)

multiply this by r:

ar + ar^2 + ar^3 + ... + ar^n = rs (2)

Now you can see that most terms are contained in both sums, so you can subtract equation (1) from (2) and obtain:

ar^n - a = rs - s
a (r^n - 1) = (r - 1) s

s(n) = a (r^n - 1)/(r - 1)

If r < 1 you write it normally in this way:

s(n) = a (1 - r^n)/(1 - r)

note that r^n becomes smaller and smaller with increasing n, so in iii) you will get that s(100) is slightly less than a/(1 - r) = 1/0.5 = 2
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>>51122
>note that r^n becomes smaller and smaller with increasing n

* if r < 1