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Use differences to find a pattern in the sequence. 4,4,9,20,38,64,99 Assuming that the pattern​ continues, the eighth term should be?

 

Use differences to find a pattern in the sequence. 6,9,20,39,66,101,144 Assuming that the pattern​ continues, the eighth term should be?

 Sep 6, 2017
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1)   Use differences to find a pattern in the sequence. 4,4,9,20,38,64,99 Assuming that the pattern continues, the eighth term should be?

 

If you look at the difference between the terms, you get the following:

0, 5, 11, 18, 26, 35......

Now, see the difference between these new terms and you get:

5, 6, 7, 8, 9.......etc. It, therefore, follows that you will add: 10, 11, 12, 13......etc. to the FIRST differences: 35+10 =45,  45+11=56, 56+12=68.........etc. Therefore, your sequence will be:

4,4,9,20,38,64,99, 144, 200,  268........etc.

P.S. You could use this more complicated formula to generate each term:

a(n) = 1/6 (n^3 + 9n^2 - 34n + 48).

 

 

2)  Use differences to find a pattern in the sequence. 6,9,20,39,66,101,144 Assuming that the pattern continues, the eighth term should be?

 

Do the same for this one and you will see these differences:

3, 11, 19, 27, 35, 43..............

Now look at the differences between these and you get a CONSTANT of 8. So, you will have:

43+8=51,  51+8=59, 59+8=67.........etc. So your sequence will be:

 6,9,20,39,66,101,144, 195, 254, 321.........etc. 

P.S. You could also use this more complicated formula to generate each term:

a(n) = 4n^2 - 9n + 11 

 Sep 6, 2017
edited by Guest  Sep 6, 2017
 #2
avatar+26364 
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Use differences to find a pattern in the sequence
4,4,9,20,38,64,99, ...
Assuming that the pattern? continues,
the eighth term should be?

 

\(\small{ \begin{array}{lrrrrrrrrrrrrrrrrr} & 4 && 4 && 9 && 20 && 38 && 64 && 99 && {\color{red} 144} && \cdots \\ \text{1. Difference } && 0 && 5 && 11 && 18 && 26 && 35 && {\color{red} 45} && \cdots \\ \text{2. Difference } &&& 5 && 6 && 7 && 8 && 9 && {\color{red} 10} && \cdots \\ \text{3. Difference } &&&& 1 && 1 && 1 && 1 && {\color{red} 1} && \cdots \\ \end{array} } \)

 

\(\begin{array}{|rcrcr|} \hline 9 &+& {\color{red}1} &=& {\color{red}10} \\ 35&+&{\color{red}10} &=& {\color{red}45} \\ 99&+&{\color{red}45} &=& {\color{red}144} \\ \hline \end{array} \)

 

The eighth term is 99 + 35 + 9 + 1 = 144

 

laugh

 Sep 6, 2017

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