# Compound interest with an equal monthly investment

The calculation of the accumulated amount with the monthly investiment

Сomply with request of user frouzen, who asked to make /571/ - calculating accreted amount when using compound interest and additional monthly investment. The calculation of interest is also expected to be monthly (most favorable case).

In order to do not distract the user from the calculator itself it's located below. Also there is a little bit of theory and formulas for those who need it.

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### Compound interest with an equal monthly investment

Digits after the decimal point: 2
Accreted amount
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The formula of compound interest, accruing a several times during the year is $S=P(1 + \frac{j}{m})^{mn}$, where m in our case is 12 and n - is a deposit period in years.

That's the simplest case when you make a contribution immediately and without further investment to it.

Now for the more complicated case - deposit replenishment with equal monthly installments..
Note that the factor degree mn nothing more than the number of periods of interest accrual.

Thus, for the very first deposit the accreted amount for the first few years will be the same.
$S_1=P(1 + \frac{j}{m})^{mn}$
For the deposit that has been made at the end of the first month, the amount of periods of interest accrual is one fewer and the formula would look like this
$S_2=P(1 + \frac{j}{m})^{mn-1}$,
for the third deposit - like this
$S_3=P(1 + \frac{j}{m})^{mn-2}$,
...
and for the last deposit, i.e. made in the las month before the end of the term - like this
$S_{mn}=P(1 + \frac{j}{m})$,

General result is a sum of all these expressions. And there are similarity in these expressions - they are terms of geometric progression in which the first term is equal to
$P(1 + \frac{j}{m})$ and the common ratio of a geometric progression is $1 + \frac{j}{m}$.

About the geometric progression see Geometric progression

Thus,the required amount of the sum of the geometric progression formula is
$S=\frac{a_nq-a_1}{q-1}=\frac{P(1 + \frac{j}{m})^{mn}(1 + \frac{j}{m})-P(1 + \frac{j}{m})}{\frac{j}{m}}$

That's all for today

Update

Added the ability to specify individual size of the first installment(according to the user's request).
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