Find the nth term of the given sequence

In summary, the conversation discusses different approaches to finding the nth term of a sequence. One person was able to solve it using trial and error and noticed that the sequence was the cubes plus one. Another person is seeking an alternative method and is posting their working. It is mentioned that there is no unique solution and that there may be an infinite number of polynomial solutions. An example of a possible solution is given as n^3 + 1 + p(n)(n - 1)(n-2)(n-3)(n-4)(n-5) for any polynomial p.
  • #1
chwala
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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
 
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  • #2
chwala said:
I am seeking an alternative method...cheers
I guess you were supposed to notice that the sequence was the cubes plus one.
 
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  • #3
PeroK said:
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.
 
  • #4
chwala said:
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.
 
  • #5
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!:cool:
 
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  • #6
PeroK said:
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?
 
  • #7
chwala said:
Are you implying that we have other solutions for this sequence?
There must be an infinite number of polynomial solutions.
 
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  • #8
PeroK said:
There must be an infinite number of polynomial solutions.

For example [itex]n^3 + 1 + p(n)(n - 1)(n-2)(n-3)(n-4)(n-5)[/itex] for any polynomial [itex]p[/itex].
 
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