January 4th, 2017, 09:41 AM  #1 
Newbie Joined: Jan 2017 From: Portland Posts: 6 Thanks: 0  Request for Assistance  financial math
I was told the formula for determining the aggregate of 40 annual payments growing at a compounded rate of 5% is: [Y1 amount]*(1.05)^40. Using that formula (with the Y1 amount being $1,400) would give a figure of . . . what? Thanks in advance . . . not sure this is trigonometry! 
January 4th, 2017, 09:48 AM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 893 Thanks: 479 
it's not even remotely trigonometry just plug Y1 into your formula and punch some buttons on your calculator. $(1400)\times (1.05)^{40}$ 
January 4th, 2017, 10:03 AM  #3  
Math Team Joined: Jul 2011 From: Texas Posts: 2,279 Thanks: 1109  Quote:
Future value of periodic payments ... variables defined at the online calculator link ... Future Value of Periodic Payments Calculator  High accuracy calculation Last edited by skeeter; January 4th, 2017 at 10:30 AM.  
January 4th, 2017, 02:43 PM  #4 
Newbie Joined: Jan 2017 From: Portland Posts: 6 Thanks: 0 
That's quite helpful, skeeter!

January 4th, 2017, 05:08 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,424 Thanks: 569 
Your post is unclear. Are you looking for the future value of an annual payment of 1,400? If so: FV = 1400[(1.05)^40  1] / .05 = ~169119.68 Bankstatementwise, it'll look like diss: Code: YEAR DEPOSIT INTEREST BALANCE 0 .00 1 1400.00 .00 1400.00 2 1400.00 70.00 2870.00 3 1400.00 143.50 4413.50 .... 39 1400.00 7539.67 159733.03 40 1400.00 7986.65 169119.68 
January 5th, 2017, 09:36 AM  #6 
Newbie Joined: Jan 2017 From: Portland Posts: 6 Thanks: 0 
What I am looking for is the sum/aggregate of 40 annual amounts. The first year's amount is \$1,400, and the next 39 payments will be of amounts that are 5% increases over the prior year's amount. In other words, the second year's amount would be (\$1,400 + 5% growth), and the third year amount would be that amount plus 5% growth. It's a sum of figures assuming a compound growth rate of 5% over 40 years. I think the calculator skeeter posted a link to: Future Value of Periodic Payments Calculator  High accuracy calculation answers that question. Last edited by skipjack; January 5th, 2017 at 05:08 PM. 
January 5th, 2017, 09:54 AM  #7 
Newbie Joined: Jan 2017 From: Portland Posts: 6 Thanks: 0 
Thanks Denis. My post was unclear. I'm just looking for the sum of forty different figures. The first is 1,400. The second is 1,470 (1,400 plus 5%). The third is (1,470 x 1.05). And so on . . .. Each of the 40 figures will be 5% larger than the previous one. Thx. 
January 5th, 2017, 11:11 AM  #8  
Math Team Joined: Jul 2011 From: Texas Posts: 2,279 Thanks: 1109  Quote:
$\displaystyle 1400 + 1400(1.05) + 1400(1.05)^2 + \, ... \, + 1400(1.05)^{39} = 1400 \sum_{n=0}^{39} (1.05)^n = 169119.68$  
January 5th, 2017, 11:13 AM  #9 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,424 Thanks: 569 
Well Ferd, was that problem given to you in math class? If you LOOK at the representation I gave you, you'll see 1400 + 0 = 1400 1400 + 70= 1470 1400 + 143.50 = 1543.50 ....and so on... If you did not recognize that as what you're after, then ya better have a talk with your teacher. Edit...ahhh, I see Skeeter agrees! Thanks buddy!! Hope Canada beats USA (World hockey Jr.'s) tonight!! Last edited by Denis; January 5th, 2017 at 11:17 AM. 
January 5th, 2017, 11:24 AM  #10 
Newbie Joined: Jan 2017 From: Portland Posts: 6 Thanks: 0 
Thanks skeeter. Does it make sense that the ~169,111 value was derived using BOTH the "a finite geometric series with common ratio" method AND the method that you linked to yesterday ("Future value of periodic payments"  Future Value of Periodic Payments Calculator  High accuracy calculation). The value Denis derived also was ~169,111. I like the consistency. Maybe these are the same formula and it just goes by different names? Thx. 

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