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November 25th, 2016, 08:41 AM  #1 
Newbie Joined: Nov 2016 From: USA Posts: 5 Thanks: 0  need help with progression
Looking at the following progression which is dividing the first number (100) into two parts 100(.7) and 100(.3)(.9) Then each number is also divided the same way (e.g.  70(.7) and 70(.3)(.9) I am then looking for a way to sum all of the numbers. 100 70 27 49 18.9 18.9 7.29 34.3 13.23 13.23 5.103 13.23 5.103 5.103 1.9683 
November 25th, 2016, 08:56 AM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timeywimey stuff. 
Please do not double post. Dan 
November 25th, 2016, 08:58 AM  #3 
Newbie Joined: Nov 2016 From: USA Posts: 5 Thanks: 0 
I was really not sure what kind of math it is.

November 25th, 2016, 10:07 AM  #4 
Newbie Joined: Nov 2016 From: USA Posts: 5 Thanks: 0 
Does anyone have any ideas?

November 25th, 2016, 10:57 AM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond 
$100\cdot(0.97)^n$

November 25th, 2016, 02:10 PM  #6 
Newbie Joined: Nov 2016 From: USA Posts: 5 Thanks: 0 
How did you arrive at the .97?

November 25th, 2016, 02:24 PM  #7 
Newbie Joined: Nov 2016 From: USA Posts: 5 Thanks: 0 
Never mind, Is it 1  (.7  .3(.9)) 
November 25th, 2016, 07:00 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,464 Thanks: 2038 
No. The sum of the numbers in the $n$th row (starting with $n$ = 1) is $100(0.97)^{{\large n}  1}$. The corresponding sum of the numbers in all $n$ rows is $10000(1  0.97^{\large n})/3$. The 0.97 is the value of 0.7 + 0.3(0.9). 

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