# Compound Interest Formula adding annual contributions

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.

• 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... Mar 15, 2016 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. Mar 15, 2016 at 13:08

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.
• The number I get is close, but smaller than the one on this site bankrate.com/calculators/savings/… Mar 15, 2016 at 17:30
• This is what I got from entering some numbers. i.imgur.com/8Y7WfEa.jpg?1 Mar 15, 2016 at 17:40
• That's because the web calculator considers the additional payments made at the beginning of each contribution period not at the end as you stated. Mar 15, 2016 at 18:44
• How would be the formula if the additional contributions were made at the beginning of each year?. Thanks. Mar 15, 2016 at 18:50
• I added it in my answer (last equation ) Mar 15, 2016 at 18:51

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

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

• Indeed, I can continue from the F3 function on, but what is the shorter expression of final formula?. Mar 15, 2016 at 17:34
• I amended the question to include the final answer... it is the same as @alexjo's Mar 15, 2016 at 18:02
• Thanks. However, I don't seem to get the same result when using this bank rate calculator bankrate.com/calculators/savings/… Mar 15, 2016 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. Jul 21, 2023 at 20:26