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# A sequence

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482
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Please give 3 more terms of the following sequence and the formula or method used, if possible:

1, 2/3, 2/3, 4/5, 16/15, 32/21, 16/7.......etc.

Thank you for any help.

Jul 26, 2017

#1
+1

Please give 3 more terms of the following sequence and the formula or method used, if possible:

1, 2/3, 2/3, 4/5, 16/15, 32/21, 16/7.......etc.

Thank you for any help.

It is often difficult to see a pattern in such a sequence because the fractions have been reduced to their minimums. But a closer look shows that there is a simple pattern. The pattern follows this simple expression:

2^n / n(n+1), where n=term number. Since you already have 7 terms, then the next few terms will be:

2^8 / (8*9), 2^9 / (9*10), 2^10 / (10*11).....etc=256/72, 512/90, 1024/110.....etc., which when reduced, we get this: 32/9, 256/45,  512/55.......etc.

Jul 26, 2017
#2
+21869
+1

Please give 3 more terms of the following sequence and the formula or method used, if possible:

1, 2/3, 2/3, 4/5, 16/15, 32/21, 16/7.......etc.

Recursive Sequence:

$$\begin{array}{|rcll|} \hline a_n &=& 2^n \cdot \frac{1}{n(n+1)} \\ a_{n+1} &=& 2^{n+1} \cdot \frac{1}{(n+1)(n+2)} \\\\ \frac{a_{n+1}} {a_n} &=& \frac{ 2^{n+1} \cdot \frac{1}{(n+1)(n+2)} } { 2^n \cdot \frac{1}{n(n+1)} } \\ \frac{a_{n+1}} {a_n} &=& 2^{n+1-n} \cdot \frac{n(n+1)} {(n+1)(n+2)} \\ \frac{a_{n+1}} {a_n} &=& 2 \cdot \frac{n} {(n+2)} \\ \frac{a_{n+1}} {a_n} &=& \frac{2n} {(n+2)} \\\\ \mathbf{ a_{n+1} }& \mathbf{=} & \mathbf{ a_n\cdot \frac{2n} {(n+2)} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline a_7 &=& \frac{16}{7} \\\\ a_8 &=& \frac{16}{7} \cdot \frac{2\cdot 7} {(7+2)} \\ &=& \frac{16}{7} \cdot \frac{2\cdot 7} {9} \\ &=& \frac{16\cdot 2}{9} \\ & \mathbf{=} & \mathbf{ \frac{32}{9} } \\\\ a_9 &=& \frac{32}{9} \cdot \frac{2\cdot 8} {(8+2)} \\ &=& \frac{32}{9} \cdot \frac{16} {10} \\ &=& \frac{16\cdot 16}{9\cdot 5} \\ & \mathbf{=} & \mathbf{ \frac{256}{45} } \\\\ a_{10} &=& \frac{256}{45} \cdot \frac{2\cdot 9} {(9+2)} \\ &=& \frac{256}{5\cdot 9} \cdot \frac{2\cdot 9} {11} \\ &=& \frac{256\cdot 2}{5\cdot 11} \\ & \mathbf{=} & \mathbf{ \frac{512}{55} } \\ \hline \end{array}$$

Jul 27, 2017