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# Ahh sorry for all these posts but everyone who is answering me helping so much <3

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Let {a_n} is a 3rd order arithmetic sequence. The first couple of terms are: 4, 16, 42, 88. Please determine the general formula of {a_n}

Aug 13, 2018

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I think this is what you are  looking for, but I'm not sure... I'll give it my  best shot  using something known as  the "sum of differences "

4  16    42    88     160

12   26   46   72

14   20   26

6     6

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Using the polynomial  form  p(x)  = ax^3 + bx^2 + cx + d , we can  solve this system  of equations  for a,b, c and  d

a  +  b +  c +  d  =  4         [ for x  = 1, the first term]

8a + 4b + 2c + d  = 16        [for x  = 2, the second term]

27a + 9b + 3c + d  = 42      [for  x  = 3, the third term]

64a + 16b + 4c + d  = 88     [for x  = 4, the fourth term ]

The solving process is a little lengthy [ but not difficult ] ...I used WolframAlpha to find the solutions

a = 1, b  = 1,  c  = 2  and  d  = 0

So....the generating polynomial p(x) =   [ an ]  =    x^3 +  x^2 + 2x

Note that the first  five  terms are as above .....https://www.wolframalpha.com/input/?i=x%5E3+%2B+x%5E2+%2B+2x,++for++x++%3D+1,2,+3,+4,5

I hope that helped....if this is not what you want.....check back.....our member "heureka"  is  pretty knowledgeable about this sort of thing

Aug 13, 2018
edited by CPhill  Aug 13, 2018
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yes, this is exactly what I'm looking for, tysm

yasbib555  Aug 13, 2018
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Let {a_n} is a 3rd order arithmetic sequence. The first couple of terms are: 4, 16, 42, 88.
Please determine the general formula of {a_n}

$$\begin{array}{lrrrrrrrrrr} & {\color{red}d_0 = 4} && 16 && 42 && 88 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 12} && 26 && 46 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 14} && 20 && \cdots \\ \text{3. Difference } &&&& {\color{red}d_3 = 6} && \cdots \\ \end{array}$$

$$\begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } + \binom{n-1}{3}\cdot {\color{red}d_3 } \\\\ a_n &=& \binom{n-1}{0}\cdot {\color{red} 4 } + \binom{n-1}{1}\cdot {\color{red}12} + \binom{n-1}{2}\cdot {\color{red}14} + \binom{n-1}{3}\cdot {\color{red} 6} \\\\ \hline \binom{n-1}{0} &=& 1 \\ \binom{n-1}{1} &=& n-1 \\ \binom{n-1}{2} &=& \left( \frac{n-1}{2} \right) \cdot \left( \frac{n-2}{1} \right) \\ \binom{n-1}{3} &=& \left( \frac{n-1}{3} \right) \cdot \left(\frac{n-2}{2} \right) \cdot \left( \frac{n-3}{1} \right) \\ \hline \\ a_n &=& {\color{red}4 } + (n-1)\cdot {\color{red}12} + \left( \frac{n-1}{2} \right) \cdot \left( \frac{n-2}{1} \right)\cdot {\color{red}14} + \left( \frac{n-1}{3} \right) \cdot \left(\frac{n-2}{2} \right) \cdot \left( \frac{n-3}{1} \right)\cdot {\color{red}6} \\\\ &=& 4 + (n-1)\cdot 12 +(n-1)(n-2) \cdot 7 + (n-1)(n-2)(n-3) \\\\ &=& 4 + 12n-12 + (7n-7)(n-2) + (n-1)(n-2)(n-3) \\\\ &=& -8 + 12n + 7n^2-21n +14 + (n-1)(n-2)(n-3) \\\\ &=& 6 - 9n + 7n^2 + (n-1)(n-2)(n-3) \\\\ &=& 6 - 9n + 7n^2 + (n-1)(n^2-5n+6) \\\\ &=& 6 - 9n + 7n^2 + n^3-5n^2+6n-n^2+5n-6 \\\\ \mathbf{a_n} &=& \mathbf{2n + n^2 + n^3} \\ \end{array}$$

Aug 14, 2018
edited by heureka  Aug 14, 2018