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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: monkeywantsabanana on April 19, 2011, 04:18:15 pm

Title: Sequences and Series
Post by: monkeywantsabanana on April 19, 2011, 04:18:15 pm: Hey guys i'll got a few questions with this topic :(

First:

If i were given a sequence for say

1, 2, 4, 8

How would i approach this as to finding the possible rule for $t_n$ in terns of $n$ and the difference equation? Any thoughts?

I'm still trying to get my head around this. It seems so unnatural to me.

Any help would be appreciated.
Title: Re: Sequences and Series
Post by: luken93 on April 19, 2011, 04:26:04 pm: Well firstly it's geometric, as it doesn't go up by a constant amount

To find the difference, use $\frac{t_n}{t_{n-1}}$

So, using one of any of the combinations, $\frac{8}{4} = 2$

Also, you know that the first term is (n = 1) is 1

Thus, the sequence is $t_n = a \times r^{n - 1}$

$t_n = 1 \times 2^{n - 1}$

To check it's right:

$t_1 = 1 \times 2^{0}$
$t_1 = 1$

$t_2 = 1 \times 2^{1}$
$t_1 = 2$

etc...
Title: Re: Sequences and Series
Post by: monkeywantsabanana on April 19, 2011, 04:35:48 pm: Geometric series?

I have to towards the chapter.

I'm using the Essential book - 5A.

Why do they ask us the question BEFORE they introduce the concept? :O

Is there a specific type for this one?

-1, 8, -27, 64 ?

And thanks luken. (Y)
Title: Re: Sequences and Series
Post by: luken93 on April 19, 2011, 04:48:36 pm: hahaha they like to do that for that chapter if I remember...

Anyway, you have the two options, geometric series and arithmetic series. Arithmetic goes up/down by a constant difference for each term, but geometric goes up according to ^(n-1), hence the difference isn't arithmetic, but rather geometric...
After you do a few questions, it'll get much easier and a lot more understandable.

For that question, what do you see.

Firstly, they change signs every second term. A possible way to make this occur is if you do the following : (-1)^n, as every second term is positive, which is the same as -1 x -1 = +1

Also, looking at the numbers, you can see that they are cubics, (-1)^3, (2)^-3, (-3)^3 etc

So we can now set up the equation:

$t_n = a \times r^{n-1}$

$t_n = (-1)^n \times n^{3}$

To check it's correct:

$t_1 = (-1)^1 \times 1^{3}$
$t_1 = -1$

$t_2 = (-1)^2 \times 2^{3}$
$t_2 = 8$

As I said, it'll come with time, but you will soon be able to spot the trends very quickly..
Title: Re: Sequences and Series
Post by: monkeywantsabanana on April 19, 2011, 05:11:26 pm: Haha thanks a lot !

That was much more helpful than my teacher.

Cheers.
Title: Re: Sequences and Series
Post by: monkeywantsabanana on April 19, 2011, 05:14:26 pm: Oh hey can you also explain the "difference equation" for the arithmetic and geometric series please?

'cus what you just taught me was only the rule for $t_n$ in terms of $n$ right?
Title: Re: Sequences and Series
Post by: monkeywantsabanana on April 19, 2011, 05:28:05 pm: For example this one:

$1,1/4, 1/9, 1/16$

I managed to find out the rule for $t_n$ in terms of $n$ (yay!)

but I'm not sure how to find the difference equation.

Help? :o :o
Title: Re: Sequences and Series
Post by: luken93 on April 19, 2011, 05:39:56 pm: The difference equation is just r

As I showed before, it's the tn / t(n-1), but from memory I can't really remember how to apply it when it's like your example above, although you can clearly see its (1/n)^2...
Title: Re: Sequences and Series
Post by: bblovee on June 26, 2011, 09:57:39 pm: if you apply tn/t(n-1) for a few successive terms, you see that each next term is a certain fraction of the previous term
eg for 1, 1/4, 1/9, 1/16:

t2/t1 = 1/4
t3/t2 = 4/9
t3/t4 = 9/16
t5/t4 = 16/25
starting to see a pattern here?
The 5th term is t5 = 1/25, and it is 16/25 of a fraction of the previous term.
Now to express t5 in terms of t4, we have 1/25 = (4^2 / 5^2) * 1/16
so t5 = (4^2/5^2 ) * (t4-1)

you can see that tn = ((n-1)^2/n^2) * (tn-1)
or tn = ( (n-1)/n)^2 * (tn-1)