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If I have an investment that grows 10% per year, and every year I add an amount, is there a formula that can be used to find my ending balance after n years?

The following demonstrates this:

year 0: B = s*1.1^0 + a*1.1^0 = s + a
year 1: B = s*1.1 + a*1.1 + a
year 2: B = s*1.1^2 + a*1.1^2 + a*1.1 + a
year 3: B = s*1.1^3 + a*1.1^3 + a*1.1^2 + a*1.1 + a
year n: B = s*1.1^n + a*1.1^n + a*1.1^(n-1)... + a

where
    B is the balance at the start of the year
    s is the starting amount
    a is the amount added each year
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2 Answers 2

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You could use

B = ((a + a*r + r*s)*(1 + r)^n - a)/r

For example, the OP's calculation with s = 1000 and a = 100

year 3: B = s*1.1^3 + a*1.1^3 + a*1.1^2 + a*1.1 + a = 1795.10

Using the formula arrives at the same result

n = 3
r = 0.10

B = ((a + a*r + r*s)*(1 + r)^n - a)/r = 1795.10

Derivation

The OP requires the sum of the compounding amounts, plus s with interest, plus a

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Replacing the summation with the closed-form expression

from Wikipedia Geometric series formulae (which confusingly also uses r)

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From Investopaedia

Compound interest = [S (1 + i) ^ n] – S

  • S = Principal
  • n = number of years
  • i = interest rate

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