Hi good people!,

I am quite confident with Arithmetic and Geometric sequences, however this one has me. They say we have to calculate the formula for the nth term, using this sequence:

\({1 \over 5} ; {3 \over 8};{9 \over 13};{27 \over 20}\)

The answer is:

\(Tn = {1*{3^{n-1}} \over n^2+4}\)

Please help me with this. Thank you kindly..

juriemagic Apr 12, 2018

#1**+1 **

What is it that you don't understand? You are given the formula to calculate any term of the sequence: T(n) =1*3^(n-1) / n^2+4. All you have to do is substitute the "term number" for n in your formula! So, first term, or 1, goes like this: T(1)=1*3^(1-1) / 1^2+4 =1/5. For the 2nd term, or 2, you get this: T(2) =1*3^(2-1) / 2^2+4=3/8. Third term, or 3, you have: T(3) =1*3^(3-1) / 3^2+4=9/13. The 4th term, or 4, would be:T(4)=1*3^(4-1) / 4^2+4 =27/20........ and so on. So, the 10th term, or 10, would be:

T(10) =1*3^(10 - 1) / 10^2 + 4 =19,683 / 104 ..........etc. Do you get it now?.

So, to derive their formula, the numerator is 3 raised to to the power of term number -1, or 3^n - 1. The denominator is the term number, n squared + 4.

Guest Apr 12, 2018

edited by
Guest
Apr 12, 2018

#2**+1 **

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.

juriemagic
Apr 12, 2018

#3**+1 **

Hi Juri: But it is quite clear how to derive it from the sequence as I edited it in the last 2 lines.

Guest Apr 12, 2018

#4**+4 **

No Guest, you have not interpreted Juriemagic's question properly.

He has the answer but he wants to know how he could have seen it, or worked it out by himself.

Lets look at the numerators seperately from the denominators

numerator - I notice straight off that these are powers of 3

n | 1 | 2 | 3 | 4 | n |

T_n | 1 | 3 | 9 | 27 | |

3^(1-1) | 3^(2-1) | 3^(3-1) | 3^(4-1) | 3^(n-1) |

denominator

5 8 13 20

now I am going to subtract the preceding term from each of these.

8-5=3 13-8=5 20-13=7

3 5 7 If these had all been the same then it could have been expresed as polynomial of degree 1

Now I wll subtract again

5-3=2 7-5=2

2 2 These are the same so I can express this sequence as a polynomial of degree 2

n | 1 | 2 | 3 | 4 | n |

n^2 | 1 | 4 | 9 | 16 | n^2 |

t_n | 5 | 8 | 13 | 20 | |

t^n+4 | 5 | 8 | 13 | 20 |

So putting this altogether I have found the sequence to be.

\(T_n=\dfrac{3^{n-1}}{n^2+4}\)

If you have any questions Juriemagic just ask. If you want to be sure that I see it, send me a personal message and include the address of the question. :)

Melody Apr 12, 2018

#5**+2 **

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...

juriemagic
Apr 12, 2018

#6**+2 **

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!!

juriemagic
Apr 12, 2018

#7**+2 **

Hi Juriemagic,

That is good except I am not sure about this bit.

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

Where did you get 2a=2 from?

I did that subtraction **twice** and then had all the numbers the same

(all 2 in this case, but the 2 is not relevent, it is that they are **all the same on the second subtraction** that is relevant)

So I knew I could express this a polynomial of degree 2.

After that I just looked at n^2 that is 1, 4, 9, 16

and compared it to the sequences 5, 8, 13, 20

and I could just see that I had to add 4. Denominator = n^2+4

[what you did might be right, but I don't understand it. ]

Melody Apr 12, 2018

#8**+1 **

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

\(a+b+c=T_1\)

juriemagic
Apr 16, 2018