MHB Determine the general term for a sequence

[Amer]

What is the general term for the sequence

8,12,18,27,...

First of all I know that i can make a polynomial or whatever function pass through these points but I make a relation I just want to build the general term of it
I took the difference between any two terms
I choose 40
8 , 12 , 18 , 27 , 40 , 58.
4 , 6 , 9 ,13 , 18
2 , 3 ,4 , 5
1 , 1 , 1
so first row is constant second is n+1
but third how I can make it .

thanks

 

[Dustinsfl]

Re: Determine the general term for a sequnce

Since it is 3 rows down, you are looking for a polynomial of degree n = 3, i.e $an^3 + bn^2 + cn + d$

We know that
$$
\begin{alignat*}{3}
a + b + c + d & = & 8\\
8a + 4b + 2c + d & = & 12\\
27a + 9b + 3c + d & = & 18\\
64a + 16b + 4c + d & = & 27
\end{alignat*}
$$
You have a system of 4 equations with 4 unknowns. What should you do now?

 

[QuestForInsight]

Re: Determine the general term for a sequnce

This also works: $\displaystyle f(k) = \frac{3^k}{2^{k-3}} ~~ (k \ge 0)$.

 

[Sudharaka]

Re: Determine the general term for a sequnce

This also works: $\displaystyle f(k) = \frac{3^k}{2^{k-3}} ~~ (k \ge 0)$.

Yeah, it seems to work, can you tell us how you obtained it? :)

 

[QuestForInsight]

Re: Determine the general term for a sequnce

Yeah, it seems to work, can you tell us how you obtained it? :)

I've noticed that $2(8)-\frac{1}{2}(8) = 12$, $2(12)-\frac{1}{2}(12) = 18$, $2(18)-\frac{1}{2}(18) = 27.$

We wish to find the sequence that satisfies $a_{k+1} = \frac{3}{2}a_{k} ~~ (k \ge 0; ~ a_{0} = 8)$.

Define the generating function $\displaystyle A(x) = \sum_{k \ge 0}a_{k}x^k$ (our aim is to find $a_{k}$).

Multiplying the left side of our sequence by $x^k$ and summing it over all $k$ we have:

$\displaystyle \sum_{k \ge 0} a_{k+1}x^k = \sum_{k \ge 1}a_{k}x^{k-1} = \sum_{k \ge 0}a_{k}x^{k-1}-\frac{a_{0}}{x} = \frac{1}{x}\sum_{k\ge 0}a_{k}x^k-\frac{8}{x} = \frac{A(x)}{x}-\frac{8}{x}.$

Multiplying the right side of our sequence by $x^k$ and summing it over all $k$ we have:

$\displaystyle \frac{3}{2}\sum_{k \ge 0} a_{k}x^k = \frac{3}{2}A(x).$ Thus $\displaystyle \frac{A(x)}{x}-\frac{8}{x} = \frac{3}{2}A(x) \implies A(x) = \frac{16}{2-3x}$.

Expanding this $A(x)$ in power series we find that it's $\displaystyle A(x) = \sum_{k \ge 0}\frac{3^k}{2^{k-3}}x^{k}$.

Thus $ \displaystyle a_{k} =\frac{3^k}{2^{k-3}} ~~~ (k \ge 0).$

 

[MarkFL]

Re: Determine the general term for a sequnce

This also works: $\displaystyle f(k) = \frac{3^k}{2^{k-3}} ~~ (k \ge 0)$.

$\displaystyle f(4)=40.5,\,f(5)=60.75$

It works only for $\displaystyle 0\le k\le3$

 

[QuestForInsight]

Re: Determine the general term for a sequnce

$\displaystyle f(4)=40.5,\,f(5)=60.75$

It works only for $\displaystyle 0\le k\le3$

I think you misread the thread.

 

[Dustinsfl]

Re: Determine the general term for a sequnce

I think you misread the thread.

The 4th term is stated as 40 and all difference down to the 3rd row must be 1. At 4, that $f$ fails.

 

[QuestForInsight]

Re: Determine the general term for a sequnce

The 4th term is stated as 40 and all difference down to the 3rd row must be 1. At 4, that $f$ fails.

Okay sorry. Then it's me who misread the original thread. I thought all the given terms of the concerned sequence were just 8, 12, 18, and 27 (and to be fair the question does appear that way in the original post). My apologies. I should have paid more attention.

 

[MarkFL]

Re: Determine the general term for a sequnce

Okay sorry. Then it's me who misread the original thread. I thought all the given terms of the concerned sequence were just 8, 12, 18, and 27 (and to be fair the question does appear that way in the original post). My apologies. I should have paid more attention.

I can easily see it being taken either way. I still applaud your ingenuity!(Yes)

 

[Amer]

Re: Determine the general term for a sequnce

Thank you all