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August 2nd, 2018, 06:38 PM  #1 
Newbie Joined: Aug 2018 From: United States Posts: 2 Thanks: 0  algebra 2 questions! Need help ASAP! 1. Explain why we can have a converging infinite series when 0 < r < 1. Hint: Use "sum of a finite geometric series" as a basis for your explanation. 2. John begins a savings account as follows. On week 1, he puts 1 penny in his bank, on Week 2, he puts in 2 pennies, in Week 3 he puts in 3 pennies, etc. Write an equation that will model how much money he has saved after 'n' weeks and use it to find the amount of money that he will save in one year. How much money did he put in his bank during the final week? Show all work. THANK YOU! 
August 2nd, 2018, 07:38 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,712 Thanks: 1805  1. As "a convergent infinite series" could be a geometric series, it presumably suffices to consider geometric series (using the hint provided). If a geometric series has n terms and common ratio r, consider the result of multiplying the series by 1  r and hence deduce a formula for the sum. If 1 < r < 1, how does that sum behave as n increases indefinitely? 2. In weeks 1 and n, John saves an average of (1 + n)/2 pennies per week. Do you see why this weekly average is the same when weeks 1 to n are considered together? The total number of pennies saved in those n weeks is therefore n(1 + n)/2. To proceed further, note that a year is slightly longer than 52 weeks, making it slightly unclear whether you should include the pennies saved in week 53. 
August 3rd, 2018, 04:02 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,296 Thanks: 934 
n = 5: 1 + 2 + 3 + 4 + 5 = 15 n(n + 1) / 2 = 5(5 + 1) / 2 = 5(6) / 2 = 30/2 = 15 
August 7th, 2018, 10:06 AM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  Quote:
The second problem is an "arithmetic series" since each term is just the previous term plus 1. For example, if n= 4, the series is 1+ 2+ 3+ 4= 10 One way to do this is in general is to write S= 1+ 2+ 3+ ...+ (n 1)+ n and then reverse the sum to write S= n+ (n1)+ ...+ 3+ 2+ 1. Adding the two equations vertically, so that we are adding 1+ n= n+ 1, 2+ (n1)= n+ 1, etc. we see that each pair of numbers adds to n+ 1. There are n such pairs so the total on the right is n(n+ 1). We have 2S= n(n+1) so that S= n(n+1)/2. (Check that when n= 4 that is 4(4+ 1)/2= 4(5)/2= 20/2= 10 as before.) Last edited by Country Boy; August 7th, 2018 at 10:16 AM.  

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