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**mathgeek62****Member**- Registered: 2006-10-22
- Posts: 8

does anyone know how to write out the nth term? here is my problem...

1 2 3 4 5 6 n 20

.............................................

2 -1 -4 -7 -10 -13 ? ?

so how would you write out the nth term and find the 20th? please help!:D

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,647

Subtracting 3 each time.

And we know that when n=1, then the answer is 2, so:

f(n) = 5-3n

*Last edited by MathsIsFun (2006-10-22 11:07:34)*

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**mathgeek62****Member**- Registered: 2006-10-22
- Posts: 8

wait so did you get 4 becasue 4-3=1? and how would you take that formula and find the 20th term?

*Last edited by mathgeek62 (2006-10-22 11:10:09)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,647

Ha ... you caught me as I was editing my answer.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,647

f(20) = 5 - 3×20 = 5-60 = -55

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**mathgeek62****Member**- Registered: 2006-10-22
- Posts: 8

wait so how did you get 5-3n?

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,647

Once I knew that it was "-3" each time, the next thing I wanted to know was what it is when n=0

I knew it was 2 when n=1, so I went "backwards", or "+3" to get my 5

So now I know that f(0) = 5

And that for every extra "n" it is "-3", so I get the whole formula: f(n) = 5-3n

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,647

And here is the graph: Plot of 5-3n

You can see why I wanted f(0)=5, that gives me my starting point

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**mathgeek62****Member**- Registered: 2006-10-22
- Posts: 8

ok that makes sense but why do you need to find 0?? is that for every problem you do??

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**mathgeek62****Member**- Registered: 2006-10-22
- Posts: 8

ok what about this problem??

1 2 3 4 5 6

....................................

1 4 9 16 25 36

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**mathgeek62****Member**- Registered: 2006-10-22
- Posts: 8

well isn't it increasing by 1 every time?

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,647

Well, you don't have to go for f(0), but it does make it easier.

Now the next problem is simply squaring: 1[sup]2[/sup]=1, 2[sup]2[/sup]=4, etc

so f(n) = n[sup]2[/sup]

And yes, the "difference" does increase each time (by 2), which says something interesting about squaring.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Have you ever heard of a square root?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mathgeek62****Member**- Registered: 2006-10-22
- Posts: 8

ok so you just say n2 when you are squaring? and im still not sure about the first problem because i dont get how you got f(n)=5??

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,647

The "2" has to be written smaller and higher, but yes. Or you could simply say f(n) = n×n

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,647

You know how the first problem was "-3" each time?

So what happens when you go backwards (5,4,3,2,1)? The answer *increases* by 3 each time, right? Example: from 2 to 1 then answer goes from -4 to -1.

So to go from 1 to 0 means you have to add 3 to the answer. The answer for 1 was "2", so the answer for 0 must be "2+3" equals 5.

The handy thing about 0 is that you can forget about the "3n", because 3×0=0.

so f(0)=5 only works for 0 [that is why I wrote f(0), not f(n)]

For all other values you need f(n) = 5-3n

Try it out:

f(1) = 5-3 = 2

f(2) = 5 - 3×2 = 4-6 = -1

etc

Oh, and the formula does work for 0 of course: f(0) = 5 - 3×0 = 5-0 = 5

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**mathgeek62****Member**- Registered: 2006-10-22
- Posts: 8

ohhhhhh ok so let me try a problem

1 2 3 4 5

.............................................

6 12 18 24 30

so would it be 6n-12 ???

*Last edited by mathgeek62 (2006-10-22 12:23:04)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Close but no. The numbers go up by 6 each time, so there is a 6n in there, as you rightly put. However, if we go back from 1 to get f(0), then that is 6-6=0, so we don't need to add or take anything away.

The answer is simply 6n.

Why did the vector cross the road?

It wanted to be normal.

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