+278 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}
+128474 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
+26367 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}
\(\begin{array}{lrrrrrrrrrr} & {\color{red}d_0 = 4} && 16 && 42 && 88 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 12} && 26 && 46 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 14} && 20 && \cdots \\ \text{3. Difference } &&&& {\color{red}d_3 = 6} && \cdots \\ \end{array}\)
\(\begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } + \binom{n-1}{3}\cdot {\color{red}d_3 } \\\\ a_n &=& \binom{n-1}{0}\cdot {\color{red} 4 } + \binom{n-1}{1}\cdot {\color{red}12} + \binom{n-1}{2}\cdot {\color{red}14} + \binom{n-1}{3}\cdot {\color{red} 6} \\\\ \hline \binom{n-1}{0} &=& 1 \\ \binom{n-1}{1} &=& n-1 \\ \binom{n-1}{2} &=& \left( \frac{n-1}{2} \right) \cdot \left( \frac{n-2}{1} \right) \\ \binom{n-1}{3} &=& \left( \frac{n-1}{3} \right) \cdot \left(\frac{n-2}{2} \right) \cdot \left( \frac{n-3}{1} \right) \\ \hline \\ a_n &=& {\color{red}4 } + (n-1)\cdot {\color{red}12} + \left( \frac{n-1}{2} \right) \cdot \left( \frac{n-2}{1} \right)\cdot {\color{red}14} + \left( \frac{n-1}{3} \right) \cdot \left(\frac{n-2}{2} \right) \cdot \left( \frac{n-3}{1} \right)\cdot {\color{red}6} \\\\ &=& 4 + (n-1)\cdot 12 +(n-1)(n-2) \cdot 7 + (n-1)(n-2)(n-3) \\\\ &=& 4 + 12n-12 + (7n-7)(n-2) + (n-1)(n-2)(n-3) \\\\ &=& -8 + 12n + 7n^2-21n +14 + (n-1)(n-2)(n-3) \\\\ &=& 6 - 9n + 7n^2 + (n-1)(n-2)(n-3) \\\\ &=& 6 - 9n + 7n^2 + (n-1)(n^2-5n+6) \\\\ &=& 6 - 9n + 7n^2 + n^3-5n^2+6n-n^2+5n-6 \\\\ \mathbf{a_n} &=& \mathbf{2n + n^2 + n^3} \\ \end{array} \)
