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 January 23rd, 2017, 09:56 AM #1 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,115 Thanks: 913 Financial formula To de-mistify(!) 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: USA Posts: 2,091 Thanks: 1087 $f =\dfrac d i \displaystyle{\sum_{k=1}^n}~u^{k m}$ $f = \dfrac d i \displaystyle{\sum_{k=1}^n}~\left(u^m\right)^k =$ $u^m \dfrac {d}{i} \dfrac{1 - u^{nm}}{1-u^m}$
 January 23rd, 2017, 04:29 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,115 Thanks: 913 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 d = 100, i = .10, u = 1.1, m = 3, n = 3 f = d(u^(1m))/i + d(u^(2m))/i + d(u^(3m))/i = ~2460.51
January 23rd, 2017, 05:58 PM   #4
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Quote:
 Originally Posted by Denis 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 d = 100, i = .10, u = 1.1, m = 3, n = 3 f = d(u^(1m))/i + d(u^(2m))/i + d(u^(3m))/i = ~2460.51
can you write your formula in latex?

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: 13,115 Thanks: 913 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
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Quote:
 Originally Posted by Denis 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.
so this?

$f =\dfrac d i \displaystyle{\sum_{k=1}^n}~\left(u^{k m}-1\right)$

$f = \dfrac d i \left(\displaystyle{\sum_{k=1}^n}~u^{k m}-\displaystyle{\sum_{k=1}^n}~1\right)$

$f = \dfrac {d}{i} \left(u^m\dfrac{1 - u^{nm}}{1-u^m} - n\right)$

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: 13,115 Thanks: 913 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 Thanks again.
 January 26th, 2017, 05:21 AM #8 Senior Member   Joined: May 2008 Posts: 290 Thanks: 76 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(1.10^3-1)/.10=331, year end payment for 2nd year is 200(1.10^3-1)/.10=662, year end payment for 3rd year is 300(1.10^3-1)/.10=993; thus giving a common difference of 331. With R = 331 and an effective rate of i = 1.10^3-1 and n = 3 years, the problem is reduced to a blind application of the aforementioned formula of S = (R/i){(1+i)[(1+i)^n-1]/i-n} which bears some resemblance to Sir Romsek's formula. Thanks from Denis Last edited by jonah; January 26th, 2017 at 05:24 AM.

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