Is there an equation for compounding interest while adding to the principle amount supposed I invested \$5000 at 10 % per annum , but at the beginning of each year I add another \$5000. So I will have (\$5000 + \$500) + \$5000 = \$10500 at the beginning of the 2nd year and compound that at 10% and so on.
So in short suppose I invest \$5000 at 10 % per annum and add \$5000 at the beginning of each year , how much would I have after 30 years. Is there a simple equation to work that out?
P.S. I already solved it using excel but was wondering if there was a simpler formula
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:
$$
\begin{align*}
A_t &amp;= (1+r)^{t} A_0 + Q \sum_{j = 0}^{t-1} (1+ r)^j,\\
&amp;=(1 + r)\left[(1+ r)^{t-1} A_0 + Q \sum_{j = 0}^{t-2} (1+ r)^j\right] + Q,\\
&amp;= (1+ r) A_{t-1} + Q.
\end{align*}
$$
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:
$$
A_t = (1+ r)^t A_0 + Q \frac{(1+ r)^{t} - 1}{r}
$$