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May 16th, 2018, 08:37 AM  #1 
Newbie Joined: May 2018 From: italy Posts: 1 Thanks: 0  Can you solve these sequences?
Hi Can you solve these? 1) 12345, 21295, 218143, 2812793, ? 2) 110, 108, 99, 81, ? Best regards. 
May 16th, 2018, 09:24 AM  #2 
Senior Member Joined: Feb 2010 Posts: 697 Thanks: 135  
May 16th, 2018, 10:43 AM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,203 Thanks: 1157 
for #2 $\begin{align*} &100108=2\\ \\ &10899=9\\ \\ &9981=18=2\cdot 9\\ \\ &81(9\cdot 18) = 81 \end{align*}$ 
May 16th, 2018, 10:45 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,203 Thanks: 1157  
May 16th, 2018, 12:00 PM  #5 
Senior Member Joined: Feb 2010 Posts: 697 Thanks: 135  I assume you are kidding. Take (1,110), (2,10, (3,99), (4,81), (5,13.2) and fit a 4th degree polynomial. Any finite sequence of $\displaystyle n$ terms can have an $\displaystyle (n+1)^{\text{st}}$ term that can take on any value. Consider this sequence 1, 4, 9, 16, 25. Then the next term is not 36. It is 156. Use this: $\displaystyle a_n=n^2 + (n1)(n2)(n3)(n4)(n5)$ 
May 16th, 2018, 01:53 PM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,629 Thanks: 954  

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