My Math Forum need help with progression

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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: 2,256 Thanks: 926 Math Focus: Wibbly wobbly timey-wimey 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: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond $100\cdot(0.97)^n$ Thanks from topsquark
 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: 20,919 Thanks: 2201 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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