Is there an equation for compounding interest while adding to the principle amount

Let $A_t$ be the amount of money at the beginning of year $t$ and let $r$ be the interest rate. Assume that each year, we invest an additional amount of $Q$ Then we can recursively write this as: $$ A_t = (1+r)A_{t-1} + Q. $$ So the amount at time $t$, $A_t$ is the amount last year $A_t$ plus a percentage $r$ of $A_t$ plus $Q$. We would like to find a closed form solution for this problem, i.e. remove the $A_t$ from the right hand side.

Notice that, by recursion, also: $$ A_{t-1} = (1+r)A_{t-2} + Q. $$ As such substituting for $A_{t-1}$ in our original equation gives: $$ A_t = (1+r)^2 A_{t-2} + (1+r)Q + Q. $$ Next, substituting out $A_{t-2} = (1+ r) A_{t-3} + Q$ gives: $$ A_t = (1+r)^3 A_{t-3} + (1+ r)^2 Q + (1+ r) Q + Q. $$ There's a pattern here. $A_{t-j}$ gets multiplied by $(1+r)^j$ and then terms are added: $Q$, $(1+r)Q, (1+r)^2 Q, ldots, (1+ r)^{j-1} Q$.

We conjecture therefore that the solution will be of the form: $$ A_t = (1+r)^t A_0 + Q sum_{j = 0}^{t-1} (1+ r)^j $$ We can show that this is indeed the solution by induction. For $t = 1$ we have: $$ A_1 = (1+ r) A_0 + Q. $$ which is true.

Also: $$ begin{align*} A_t &= (1+r)^{t} A_0 + Q sum_{j = 0}^{t-1} (1+ r)^j, &=(1 + r)left[(1+ r)^{t-1} A_0 + Q sum_{j = 0}^{t-2} (1+ r)^jright] + Q, &= (1+ r) A_{t-1} + Q. end{align*} $$ Now, to finish up we can simplify the geometric sum: $$ sum_{j = 0}^{t-1} (1+ r)^j = frac{(1+ r)^{t}- 1}{r} $$ As such: $$ A_t = (1+ r)^t A_0 + Q frac{(1+ r)^{t} - 1}{r} $$

Let $a =$ Principle
Let $b = 1$ + interest rate
Let $c_{1} =$ future value after year $1$
Let $d =$ additional investment

$c_{1} = ab$
$c_{2} = (c_{1} + d)b$
$c_{3} = (c_{2} + d)b$
$c_{4} = (c_{3} + d)b$
$c_{5} = (c_{4} + d)b$

simplified form for $c_{2}$

$c_{2} = (c_{1} + d)b= c_{1}b + db$

simplified form for $c_{3}$

$c_{3} = (c_{2} + d)b = (c_{1}b + db + d)b = c_{1}b^{2} + db^{2} + db $

simplified form for $c_{4}$

$c_{4} = (c_{3} + d)b = (c_{1}b^{2} + db^{2} + db + d)b = c_{1}b^{3} + db^{3} + db^{2} + db$

simplified form for $c_{5}$

$c_{5} = (c_{4} + d)b = (c_{1}b^{3} + db^{3} + db^{2} + db + d)b = c_{1}b^{4} + db^{4} + db^{3} + db^{2} + db$

$therefore$

for $ ngeq 2 in mathbb{N} $

$c_{n} = c_{1}b^{n-1} + db^{n-1} + db^{n-2} + ... + db^{2} + db$

$c_{n} = c_{1}b^{n-1} + d(b^{n-1} + b^{n-2} + ... + b^{2} + b)$

$c_{n} = c_{1}b^{n-1} + d(b + b^{2} + ...+ b^{n-2} + b^{n-1})$

$c_{n} = c_{1}b^{n-1} + d[sum_{1}^{n-1}b^{x}]$

$c_{n} = (ab)b^{n-1} + d[sum_{1}^{n-1}b^{x}]$

$c_{n} = ab^{n} + d[sum_{1}^{n-1}b^{x}]$

Using summation of a power series

$c_{n} = ab^{n} + d[frac{b(1-b^{n-1})}{1-b}]$, for $ n in mathbb{N} $