# Make the Equation

• Oct 11th 2006, 08:55 PM
Lammalord
Make the Equation
okay, i cant reamber how to make the equation for this problem and i need help:

okay it starts a 0, then the first number is 3000, second, 9000, third, 18000, 4th 30000 so on..

so it increases by 3000 for each time it adds, but i cant reamber how to make the problem, that was like 2 years agao :(
• Oct 11th 2006, 11:22 PM
CaptainBlack
Quote:

Originally Posted by Lammalord
okay, i cant reamber how to make the equation for this problem and i need help:

okay it starts a 0, then the first number is 3000, second, 9000, third, 18000, 4th 30000 so on..

so it increases by 3000 for each time it adds, but i cant reamber how to make the problem, that was like 2 years agao :(

The n-th term is:

s(n) = 3000*(n-1).

alternativly you can write it as:

s(1)=0,
s(n+1) = s(n) + 3000.

RonL
• Oct 11th 2006, 11:25 PM
earboth
Quote:

Originally Posted by Lammalord
okay, i cant reamber how to make the equation for this problem and i need help:
okay it starts a 0, then the first number is 3000, second, 9000, third, 18000, 4th 30000 so on..
so it increases by 3000 for each time it adds, but i cant reamber how to make the problem, that was like 2 years agao :(

Hi,

I rewrite your problem using a table:

Code:

n      a_n        factorized
1      3000      = 1 * 3000
2      9000      = 3 * 3000
3      18000      = 6 * 3000
4      30000      = 10 * 3000
...    ......        ......

In general you need an equation : a_n = f * 3000, where f is a factor with respect to n.

As you may have noticed the factor is exactly the sum of the n:
Code:

n    sum
1      1
2      1+2 = 3
3      3+3 = 6
4      6+4 = 10
...    ...

The sum of the first n positive integers is calculated by: S = (1/2)n(n+1)

a_n = (1/2)n(n+1)*3000

EB
• Oct 12th 2006, 04:36 AM
Lammalord
thanks for the equation guys, ya its sort of sad it wasnt even math homework or anything.. ill know to come back here if i ever have a problem again
• Oct 12th 2006, 04:58 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
The n-th term is:

s(n) = 3000*(n-1).

alternativly you can write it as:

s(1)=0,
s(n+1) = s(n) + 3000.

RonL

What rubbish, next time read the question properly:

the sequence is: 0, 3000, 9000, 18000, 30000, ...

the n-th term is:

s(n)=3000*n*(n-1)/2

(differes from earboth's solution as his sequence starts with 3000, not with 0
as mine does)

RonL
• Oct 12th 2006, 05:03 AM
Lammalord
i was wondering about that, it didnt work, but it did remind me how to set up thouse type's of qustions..