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January 23rd, 2017, 09:56 AM  #1 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,676 Thanks: 697  Financial formula
To demistify(!) a bit: f = future value, d = periodic deposit, i = interest rate, u = 1+i, m = positive integer constant, n = number of periods: in other words, a complicated annuity problem! As a series, future value is: f = d(u^(1m))/i + d(u^(2m))/i + d(u^(3m))/i + ..... + d(u^(nm))/i I've been trying to work out a closed form formula: any suggestions? 
January 23rd, 2017, 01:44 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,490 Thanks: 749 
$f =\dfrac d i \displaystyle{\sum_{k=1}^n}~u^{k m}$ $f = \dfrac d i \displaystyle{\sum_{k=1}^n}~\left(u^m\right)^k =$ $u^m \dfrac {d}{i} \dfrac{1  u^{nm}}{1u^m}$ 
January 23rd, 2017, 04:29 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,676 Thanks: 697 
Don't think so, Romsek; your formula gives ~5460.51 for the example below...an EXACT 3000 too much... Code: year deposit interest balance 0 .00 1 100.00 .00 100.00 2 100.00 10.00 210.00 3 100.00 21.00 331.00 4 200.00 33.10 564.10 5 200.00 56.41 820.51 6 200.00 82.05 1,102.56 7 300.00 110.26 1,512.82 8 300.00 151.28 1,964.10 9 300.00 196.41 2,460.51 f = d(u^(1m))/i + d(u^(2m))/i + d(u^(3m))/i = ~2460.51 
January 23rd, 2017, 05:58 PM  #4  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,490 Thanks: 749  Quote:
I get 5460.51 for that using the sum formula directly and using the closed form formula  
January 23rd, 2017, 07:55 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,676 Thanks: 697 
My bad; forgot the "1"'s d=100, i = .1, u=1.1 d*(u^3  1)/i + d*(u^6  1)/i + d*(u^9  1)/i = 2460.51 What it boils down to is 3 annuities of 100 annually, one for 3 years, one for 6 years and one for 9 years. 
January 23rd, 2017, 08:38 PM  #6  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,490 Thanks: 749  Quote:
$f =\dfrac d i \displaystyle{\sum_{k=1}^n}~\left(u^{k m}1\right)$ $f = \dfrac d i \left(\displaystyle{\sum_{k=1}^n}~u^{k m}\displaystyle{\sum_{k=1}^n}~1\right)$ $f = \dfrac {d}{i} \left(u^m\dfrac{1  u^{nm}}{1u^m}  n\right)$ This gives this correct results for that set of parms you posted.  
January 24th, 2017, 07:04 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,676 Thanks: 697 
YES! Works fine. I didn't like mine, as I was converting each "period of 3" into a "period of 1", as per these contortions: y = number of years (multiple of 3) [9 in my example] j = annual rate / 100 [.10 in example] f = first deposit [$100 in example ] n = FLOOR(y/3) + 1 [4 in example] i = (1 + j)^3  1 [.331 in example] d = f(j^2 + 3j + 3) [331.00 in example] Code: d[(1 + i)^n  in  1] f =  i^2 
January 26th, 2017, 05:21 AM  #8 
Senior Member Joined: May 2008 Posts: 258 Thanks: 50  Beer soaked ramblings follow. Accumulated value of a simple increasing annuity (payments are in arithmetic progression). The given example can be viewed as an annuity whose payments are made at the end of 4 months for 3 years; year end payment for 1st year is 100(1.10^31)/.10=331, year end payment for 2nd year is 200(1.10^31)/.10=662, year end payment for 3rd year is 300(1.10^31)/.10=993; thus giving a common difference of 331. With R = 331 and an effective rate of i = 1.10^31 and n = 3 years, the problem is reduced to a blind application of the aforementioned formula of S = (R/i){(1+i)[(1+i)^n1]/in} which bears some resemblance to Sir Romsek's formula. Last edited by jonah; January 26th, 2017 at 05:24 AM. 

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