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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}

yasbib555  Aug 13, 2018
 #1
avatar+91160 
+3

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

              0

 

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

 

 

cool cool cool

CPhill  Aug 13, 2018
edited by CPhill  Aug 13, 2018
 #2
avatar+211 
+1

yes, this is exactly what I'm looking for, tysm

yasbib555  Aug 13, 2018
 #3
avatar+20153 
+2

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}

 

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

 

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

 

laugh

heureka  Aug 14, 2018
edited by heureka  Aug 14, 2018

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