# Thread: formula of a sequence ...

1. ## formula of a sequence ...

What is the formula for this sequence:

1, 4, 8, 13, 19, ...

I know that the sequence is:

$\displaystyle a_n=\frac {1}{2}(n^2+3n-2)$

The question is (what is the method of finding this answer ?).

2. ## Re: formula of a sequence ...

Originally Posted by razemsoft21
What is the formula for this sequence:

1, 4, 8, 13, 19, ...

I know that the sequence is:

$\displaystyle a_n=\frac {1}{2}(n^2+3n-2)$

The question is (what is the method of finding this answer ?).

3. ## Re: formula of a sequence ...

Originally Posted by razemsoft21
What is the formula for this sequence:

1, 4, 8, 13, 19, ...

I know that the sequence is:

$\displaystyle a_n=\frac {1}{2}(n^2+3n-2)$

The question is (what is the method of finding this answer ?).
Construct a difference table, you will find the second differences are constant, so the sequence is a quadratic in the index.

Then since a quadratic is of the form $\displaystyle an^2+bn+c$ you have:

$\displaystyle \phantom{4}a\phantom{4}+\phantom{2}b+c=1$

$\displaystyle 4a+2b+c=4$

$\displaystyle 9a+3b+c=8$

which can be solved for $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$

CB

4. ## Re: formula of a sequence ...

Originally Posted by razemsoft21
What is the formula for this sequence:

1, 4, 8, 13, 19, ...

I know that the sequence is:

$\displaystyle a_n=\frac {1}{2}(n^2+3n-2)$

The question is (what is the method of finding this answer ?).
The sequence is solution of the 'difference equation'...

$\displaystyle a_{n+1}=a_{n}+3+n\ ,\ a_{0}=1$ (1)

... which can be written as...

$\displaystyle \Delta_{n}=a_{n+1}-a_{n}= 3+n\ ,\ a_{0}=1$ (2)

The 'standard solution' of (1) is...

$\displaystyle a_{n}= a_{0}+\sum_{k=0}^{n-1} \Delta_{k}= a_{0}+ 3 n + \sum_{k=0}^{n-1} k =$

$\displaystyle = 1 + 3n + \frac{n\ (n-1)}{2}= \frac{2+5 n +n^{2}}{2}$ (3)

Regarding the solving procedure of (1) see the tutorial post on the difference equation in the section 'Discrete Mathematics'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. ## Re: formula of a sequence ...

Originally Posted by CaptainBlack
Construct a difference table, you will find the second differences are constant, so the sequence is a quadratic in the index.

Then since a quadratic is of the form $\displaystyle an^2+bn+c$ you have:

$\displaystyle \phantom{4}a\phantom{4}+\phantom{2}b+c=1$

$\displaystyle 4a+2b+c=4$

$\displaystyle 9a+3b+c=8$

which can be solved for $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$

CB
The 3 equations works, but

how can we Construct a difference table

the defference is 3,4,5,6, ...

how do we know that the sequence is a quadratic ?

6. ## Re: formula of a sequence ...

Originally Posted by razemsoft21
The 3 equations works, but

how can we Construct a difference table

the defference is 3,4,5,6, ...

how do we know that the sequence is a quadratic ?
What you have there are the first differences, when you take the differences again you get a constant, and you should know that the constant n-th differences mean the original sequence was a polynomial of degree n in the index. Hence as the second differences are constant the original sequence is a quadratic in the index.

CB

7. ## Re: formula of a sequence ...

Originally Posted by chisigma
The sequence is solution of the 'difference equation'...

$\displaystyle a_{n+1}=a_{n}+3+n\ ,\ a_{0}=1$ (1)

... which can be written as...

$\displaystyle \Delta_{n}=a_{n+1}-a_{n}= 3+n\ ,\ a_{0}=1$ (2)

The 'standard solution' of (1) is...

$\displaystyle a_{n}= a_{0}+\sum_{k=0}^{n-1} \Delta_{k}= a_{0}+ 3 n + \sum_{k=0}^{n-1} k =$

$\displaystyle = 1 + 3n + \frac{n\ (n-1)}{2}= \frac{2+5 n +n^{2}}{2}$ (3)

Regarding the solving procedure of (1) see the tutorial post on the difference equation in the section 'Discrete Mathematics'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Equation (3) does not match mine

8. ## Re: formula of a sequence ...

Originally Posted by CaptainBlack
What you have there are the first differences, when you take the differences again you get a constant, and you should know that the constant n-th differences mean the original sequence was a polynomial of degree n in the index. Hence as the second differences are constant the original sequence is a quadratic in the index.

CB
uderstood, thanks

finaly how can we form the 3 equations?

9. ## Re: formula of a sequence ...

Originally Posted by razemsoft21
Equation (3) does not match mine
I do not follow his workings either.
Here is my way.
$\displaystyle a_1=1$ and if $\displaystyle n\ge 2$ then $\displaystyle a_n=a_{n-1}+(n+1)$.
Thus
$\displaystyle \begin{gathered} a_n \hfill \\ = a_1 + \sum\limits_{k = 2}^n {\left( {k + 1} \right)} \hfill \\ = 1 + \frac{{n\left( {n + 1} \right)}}{2} - 1 + \left( {n - 1} \right) \hfill \\ = \frac{1}{2}\left( {n^2 + 3n - 2} \right) \hfill \\ \end{gathered}$

10. ## Re: formula of a sequence ...

Originally Posted by razemsoft21
uderstood, thanks

finaly how can we form the 3 equations?
Let $\displaystyle P(n)=an^2+bn+c$, then:

$\displaystyle P(1)=1$

from the first term in the sequence, and:

$\displaystyle P(2)=4$

from the 2nd term, and:

$\displaystyle P(3)=8$

from the third.

These are all the terms that need be used since the three simultaneous equations will have a unique solution (and it will be the same for three equations formed from any three distinct terms in the sequence).

Also Newton has a systematic way of doing this from the difference table that you may have covered in class.

CB

11. ## Re: formula of a sequence ...

Thanks for all of you

another thank for Soroban

the 2 ways are work and understood.

12. ## Re: formula of a sequence ...

Originally Posted by razemsoft21
Equation (3) does not match mine
It is a question of 'initial index'... if the initial index is 0, then the difference equation is...

$\displaystyle a_{n+1}=a_{n}+3+n,\ a_{0}=1$ (1)

... and its solution is...

$\displaystyle a_{n}= \frac{2+5 n +n^{2}}{2}$ (2)

... if the initial index is 1, then the difference equation is...

$\displaystyle a_{n+1}=a_{n}+2+n,\ a_{1}=1$ (3)

... and its solution is...

$\displaystyle a_{n}= \frac{-2+3 n +n^{2}}{2}$ (4)

In any case You have as solution the sequence 1,4,8,13,19,...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$