juriemagic

Hi Melody,

so sorry for replying only now...when working with quadratic patterns and/or combinations of arithmetic and geometric sequences, we have a few formulas we can use:

what others call polynomial of degree 2, we refer to as either quadratic or "a second differences sequence". The same thing I know..

The 1st formula is

\(Tn=an^2+bn+c\)

Then

\(2a=d2\), so 2a is equal to the second difference.

then

\(3a+b=T_2-T_1\)

and lastly

\(a+b+c=T_1\)

Melody,

First I want to say thank you very much, you have helped me see the numerators as powers of 3.

I did some aditional study on the net, and came to this conclusion:

The numerators change to:

\(3^0;3^1;3^2;3^3\)

so this has to be :

\(3^{n-1}\) = NOMINATOR

which I understand...THEN:

you subtracted the denominators and found:

                            \(5;8;13;20\)

becomes:            \(3;5;7;9\)

which becomes:  \(2;2;2\)

From this we see that: 2a=2, therefore, a=1

                                     3a+b = 3, therefore "b" calculates to 0

                              and a+b+c=T1=5, so "c" calculates to 4

The general equation is: \(an^2+bn+c\)

So substitute everything above into this equation, and it calculates to \(n^2+4\) = DENOMINATOR

So that is then \({Nominator \over Denominator}={3^{n-1} \over n^2+4}\)

thank you kindly for your help Melody!!

MELODY!!, My Hero!!..How are you these days????..okay, heroine...

I am going to write down on a piece of paper all you said, and see if I can work my way towards the answer...will let you know if I get stuck. Thanx for the personal invite to a personal message...

No guest,

The question is not to use the formula to calculate terms, etc for example, the sequence is given and the question is: Calculate the formula for the nth Term.

the formula I gave is the ANSWER the question has. use the sequence to get to that formula as answer.

I guess she must have made a mistake, although everything still works together to the same end result...I can see she must have been explaining a "concept" to someone, as she had underlined the "-10t", "+3t" and "-t"....

anycase, I'm putting this to bed now...CPill, you were a great help!!..thanx a lot!!

Well your approach I like....I had a look and saw that you "carried" everything over to the left side in steps, and ended with:

-1 (t - 1)(t - 3) on the right side later on..

then all of that became -t^2 +4t -3..which I perfectly understand...

I was hoping to get a reasoning behind the calculation the teacher had done here;

= -10 (t - 1) - (t - 1)(t - 3) + 5 became -10t + 10 - t^2 + 3t - t - 3 + 5...

I do not understand the rule or law to change the signs like that...

I would have written:

= -10 (t - 1) - (t^2 - 3t - t + 3)..which would then become

= - 10 (t - 1) - t^2 + 3t + t - 3..which is EXACTLY what you also got..how did she get "- t" ?

I can see on the paper she used pencil to indicate she was changing the sign of the second t to a "-" inside the bracket, and the "-1" to a "+1" in this part of the sum: = -10 (t - 1) - (t - 1)(t - 3) + 5...why did she do that?...and how would this give the "-t" in any case?...I followed her calculations from thereon and everything calculates to the same end result...

CPill,

Thank you, I see it's slightly different solved than the teachers method...I'm going to study your method and will come back shortly...THANX!!

I see,

so it's a rule to take the second and third terms, should there be a "-" between them, together into a bracket, which obviously will have the "-" change to a "+"...and move on from there?..

Thanx for your time CPhill...

Hi CPhil,

yes, that is the part I understand, what I do not get is how to reverse that back to the single log term?

Hi Melody,

thanx for the reply...can you just explain how the "5" became a "4" please?