Find the nth term of the given sequence

[chwala]

Homework Statement

Find the nth term of the sequence $[2, 9 , 28, 65, 126]$

Relevant Equations

sequences

My working...i need to arrange this in an array,

$[2, 9, 28, 65, 126]$
$[ 7, 19, 37, 61 ]$
$[ 12, 18, 24 ]$
$[ 6, 6 ]$
I was able to solve this by coming up with 4 simultaneous equations,
third difference we have a constant implying a cubic sequence... therefore our sequence will be of the form $U_n= an^3 +bn^2+cn+d$
I managed to solve it...allow me to post the 4 simultaneous equations later...i finish with class...to eventually realize $U_n= n^3 +1$

I am seeking an alternative method...cheers

 

[PeroK]

I am seeking an alternative method...cheers

I guess you were supposed to notice that the sequence was the cubes plus one.

 

[chwala]

I guess you were supposed to notice that the sequence was the cubes plus one.

Yes, that's the approach i saw...trial and error method. I would want a more solid approach...i am posting my working...give me a moment.

 

[PeroK]

Yes, that's the approach i saw...trial and error method. I would want a more solid approach...i am posting my working...give me a moment.

There is no unique solution. If you look for a polynomial up to the 4th power (or higher) you should find more solutions.

 

[chwala]

ok, these are the equations;
$U_n= an^3 +bn^2+cn+d$
$2= a+b+c+d$
$9= 8a+4b+2c+d$
$28= 27a+9b+3c+d$
$65= 64a+16b+4c+d$
$126= 125a+25b+5c+d$

on working we get;
$7= 7a+3b+c$
$19= 19a+5b+c$

$12= 12a+2b$

$37= 37a+7b+c$
$61= 61a+9b+c$

$24= 24a+2b$

On Solving,
$a=1, b=0, c=0, d=1$
thus, $U_n= n^3 +1$

Finding the nth term of any quadratic sequences is a bit straightforward...one just has to note that for second difference, we use $2a=k$, where k is a constant and $a$ is the co efficient of $n^2$. cheers! bingo!

 

[chwala]

There is no unique solution. If you look for a polynomial up to the 4th power (or higher) you should find more solutions.

Are you implying that we have other solutions for this sequence?

 

[PeroK]

Are you implying that we have other solutions for this sequence?

There must be an infinite number of polynomial solutions.

 

[pasmith]

There must be an infinite number of polynomial solutions.

For example $n^3 + 1 + p(n)(n - 1)(n-2)(n-3)(n-4)(n-5)$ for any polynomial $p$.