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I'd like to know the compound interest formula for the following scenario:

P = Initial Amount i = yearly interest rate A = yearly contribution or deposit added. n = the deposits will be made for 10 consecutive years. F = final amount obtained.

I start with an initial amount and an yearly interest rate applied will be applied to it. Then, every year a contribution/deposit is made at the end of the period, that is, after the interest is applied to the previous amount. No withdrawals are made.

user3323679
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  • Have you tried simply using small values for $n$ and modelling the situation? WIth yearly interest, the interest is only applied once, along with the contribution... – Eleven-Eleven Mar 15 '16 at 12:55
  • I need the formula for a school project, that is why I made it yearly. I do know the final amount I get by using online bank calculators, but I still need to get the actual formula. – user3323679 Mar 15 '16 at 13:08

2 Answers2

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The final value $F=F'+F''$ is the sum of two components:

  • the initial deposit will produce after $n$ years at the interest rate $i$ the future value $$F'=P(1+i)^n$$
  • the periodic payments are an annuity-immediate (made at the end of each contribution period) the future value is $$ F''=A\,s_{\overline{n}|i}=A\frac{(1+i)^n-1}{i} $$ Thus, the future value $F$ is $$ F=P(1+i)^n+A\frac{(1+i)^n-1}{i}=\left(P+\frac{A}{i}\right)(1+i)^n-\frac{A}{i} $$ If the additional payments are made at the beginning of each period the annuity is an annuity due and the future value is obtained multiplying by $(1+i) $, that is $$ F''=A\frac{(1+i)^n-1}{i}(1+i) $$ and then $$ F=P(1+i)^n+A\frac{(1+i)^n-1}{i}(1+i) =\left(P+\frac{A}{d}\right)(1+i)^n-\frac{A}{d} $$ where $ d=\frac{i}{1+i} $ is the discount rate.
alexjo
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See if this helps...

So year 1; I have my initial deposit of $P$. First I need to compute the interest gained and then I have to add the new yearly contribution $A$. So by year $1$ I have

$$F_1=P+Pi+A=P(1+i)+A$$

where $F_t$ is the total in the account after $t$ years. Now that new amount will accrue yearly interest, and then you have to add $A$ to that...

$$F_2=[P(1+i)+A]+[P(1+i)+A]i+A=[P(1+i)+A](1+i)+A$$ $$=P(1+i)^2+A(1+i)+A$$

After year 3, doing the same calculation will yield for year 3

$$F_3=P(1+i)^3+A(1+i)^2+A(1+i)+A$$

$$=P(1+i)^3+A\sum_{k=0}^2(1+i)^k$$

Note that the sum is geometric and yields the formula

$$A\sum_{k=0}^2(1+i)^k=\frac{A(1+i)^3-A}{(1+i-1)}=\frac{A(1+i)^3-A}{i}$$

Thus we can extrapolate and see that for 10 years our formula is

$$F_{10}=P(1+i)^{10}+\frac{A(1+i)^{10}-A}{i}$$ which can be reduced like @alexjo did in his using annuity formulas as

$$\left(P+\frac{A}{i}\right)(1+i)^{10}-\frac{A}{i}$$

Eleven-Eleven
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    Indeed, I can continue from the F3 function on, but what is the shorter expression of final formula?. – user3323679 Mar 15 '16 at 17:34
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    I amended the question to include the final answer... it is the same as @alexjo's – Eleven-Eleven Mar 15 '16 at 18:02
  • Thanks. However, I don't seem to get the same result when using this bank rate calculator http://www.bankrate.com/calculators/savings/compound-savings-calculator-tool.aspx – user3323679 Mar 15 '16 at 18:35
  • @user3323679 This is a rather late reply but that is probably because the answers here give a formula where another yearly contribution is added to the final amount, but the website you pointed to ends the yearly contribution the year before your final year. – QuasarChaser Jul 21 '23 at 20:26