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# Use differences to find a pattern in the sequence.

0
114
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2,12,29,61,116,202,327

Feb 16, 2020

#1
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You can do this:

2,  12,  29,  61, 116,  202, 327
10,  17,   32,  55,   86,   125
7,    15,   23,   31,   39
8,      8,     8,     8

But, much better idea is to find a "closed form" for the next term, which in this case is:

a_(n-1) = 4 n^2 - 5 n + 11

a_(8-1)  = 4*7^2 - 5*7 + 11

a_(8-1)  = 172 - Then you would add this to the last term: 327 + 172 =499 which is the 8th term. And if you continue like this, your sequence will be:  2, 12, 29, 61, 116, 202, 327, 499, 726, 1016, 1377, 1817, 2344, 2966, 3691, 4527, .....etc.

Feb 16, 2020
#2
+25532
+1

Use differences to find a pattern in the sequence
$$2,\ 12,\ 29,\ 61,\ 116,\ 202,\ 327,\ \dots$$

$$\small{ \begin{array}{lccccccc} & {\color{red}d_0 = 2} && 12 && 29 && 61 && 116 && 202 && 327&& \cdots \\ \text{1. Difference } && {\color{red}d_1 = 10} && 17 && 32 && 55 && 86 && 125&& \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 7} && 15 && 23 && 31 && 39&& \cdots \\ \text{3. Difference } &&&& {\color{red}d_3 = 8} && 8 && 8&& 8&& \cdots \\ \end{array} }$$

Formula:

$$\boxed{~ \begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } + \binom{n-1}{3}\cdot {\color{red}d_3 } \end{array} ~}$$

$$\begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red} 2 } + \binom{n-1}{1}\cdot {\color{red} 10 } + \binom{n-1}{2}\cdot {\color{red} 7 } + \binom{n-1}{3}\cdot {\color{red} 8 } \\ \hline && \binom{n-1}{0} = 1 \\ && \binom{n-1}{1} = n-1 \\ && \binom{n-1}{2} = ( \frac{n-1}{2} ) \cdot ( \frac{n-2}{1} ) \\ && \binom{n-1}{3} = ( \frac{n-1}{3} ) \cdot ( \frac{n-2}{2} )\cdot ( \frac{n-3}{1} ) \\ \hline a_n &=& 1* {\color{red} 2 }+ (n-1)\cdot {\color{red}10} + ( \frac{n-1}{2} ) \cdot ( \frac{n-2}{1} )\cdot {\color{red}7} + ( \frac{n-1}{3} ) \cdot ( \frac{n-2}{2} )\cdot ( \frac{n-3}{1} )\cdot {\color{red}8} \quad | \quad \cdot 6\\ 6\cdot a_n &=& 12 + (n-1)\cdot 60 + ( n-1 ) \cdot ( n-2 )\cdot 21 + ( n-1 ) \cdot ( n-2 )\cdot ( n-3 )\cdot 8 \\ 6\cdot a_n &=& 12+ (n-1) \left[~ 60 + ( n-2 )\cdot 21 + ( n-2 )\cdot ( n-3 )\cdot 8 ~\right] \\ 6\cdot a_n &=& 12+(n-1) \left[~ 60 + 21n-42 + (n^2 - 5n + 6)\cdot 8 ~\right] \\ 6\cdot a_n &=& 12+(n-1) \left(~ 18 + 21n + 8n^2 - 40n + 48 ~\right) \\ 6\cdot a_n &=& 12+(n-1) \left(~ 66 - 19n + 8n^2 ~\right) \\ 6\cdot a_n &=& 12+ 66n - 19n^2 + 8n^3 - 66 + 19n - 8n^2 \\ 6\cdot a_n &=& -54 + 85n - 27n^2 + 8n^3 \\\\ a_n &=& \dfrac{ -54 + 85n - 27n^2 + 8n^3 }{6} \\\\ \mathbf{a_n} &=& \mathbf{ -9 + \dfrac{n(85 - 27n + 8n^2) }{6} } \\ \end{array}$$

Feb 17, 2020
edited by heureka  Feb 17, 2020