Find the nth term of the given sequence

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The sequence presented is identified as a cubic sequence, specifically of the form U_n = n^3 + 1, derived from solving four simultaneous equations. The discussion highlights that while this solution is valid, there are potentially infinite polynomial solutions to the sequence. Participants note that exploring polynomials of higher degrees may yield additional solutions. The conversation emphasizes the importance of recognizing patterns and the flexibility of polynomial equations in finding nth terms. Overall, the thread illustrates the complexity and variety of approaches to solving cubic sequences.
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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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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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.
 
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.
 
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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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?
 
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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PeroK said:
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.
 
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