<p>Let <span class="math-container">$A_t$</span> be the amount of money at the beginning of year <span class="math-container">$t$</span> and let <span class="math-container">$r$</span> be the interest rate. Assume that each year, we invest an additional amount of <span class="math-container">$Q$</span> Then we can recursively write this as:
<span class="math-container">$$
A_t = (1+r)A_{t-1} + Q.
$$</span>
So the amount at time <span class="math-container">$t$</span>, <span class="math-container">$A_t$</span> is the amount last year <span class="math-container">$A_t$</span> plus a percentage <span class="math-container">$r$</span> of <span class="math-container">$A_t$</span> plus <span class="math-container">$Q$</span>. We would like to find a closed form solution for this problem, i.e. remove the <span class="math-container">$A_t$</span> from the right hand side.</p>
<p>Notice that, by recursion, also:
<span class="math-container">$$
A_{t-1} = (1+r)A_{t-2} + Q.
$$</span>
As such substituting for <span class="math-container">$A_{t-1}$</span> in our original equation gives:
<span class="math-container">$$
A_t = (1+r)^2 A_{t-2} + (1+r)Q + Q.
$$</span>
Next, substituting out <span class="math-container">$A_{t-2} = (1+ r) A_{t-3} + Q$</span> gives:
<span class="math-container">$$
A_t = (1+r)^3 A_{t-3} + (1+ r)^2 Q + (1+ r) Q + Q.
$$</span>
There's a pattern here. <span class="math-container">$A_{t-j}$</span> gets multiplied by <span class="math-container">$(1+r)^j$</span> and then terms are added: <span class="math-container">$Q$</span>, <span class="math-container">$(1+r)Q, (1+r)^2 Q, \ldots, (1+ r)^{j-1} Q$</span>.</p>
<p>We conjecture therefore that the solution will be of the form:
<span class="math-container">$$
A_t = (1+r)^t A_0 + Q \sum_{j = 0}^{t-1} (1+ r)^j
$$</span>
We can show that this is indeed the solution by induction. For <span class="math-container">$t = 1$</span> we have:
<span class="math-container">$$
A_1 = (1+ r) A_0 + Q.
$$</span>
which is true.</p>
<p>Also:
<span class="math-container">$$
\begin{align*}
A_t &amp;= (1+r)^{t} A_0 + Q \sum_{j = 0}^{t-1} (1+ r)^j,\\
&amp;=(1 + r)\left[(1+ r)^{t-1} A_0 + Q \sum_{j = 0}^{t-2} (1+ r)^j\right] + Q,\\
&amp;= (1+ r) A_{t-1} + Q.
\end{align*}
$$</span>
Now, to finish up we can simplify the geometric sum:
<span class="math-container">$$
\sum_{j = 0}^{t-1} (1+ r)^j = \frac{(1+ r)^{t}- 1}{r}
$$</span>
As such:
<span class="math-container">$$
A_t = (1+ r)^t A_0 + Q \frac{(1+ r)^{t} - 1}{r}
$$</span></p>

CC BY-SA 4.0
2021-05-10T09:25:06.223Z
Thanks a lot , the way I ended up solving it was a tad pedestrian
Tariro Manyika
CC BY-SA 4.0
2021-05-10T09:43:16.453Z
although I have a problem with this step 

\begin{aligned}
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)^{j}\right]+Q,
\end{aligned}
Tariro Manyika
CC BY-SA 4.0
2021-05-10T09:44:06.423Z
I see what you are doing but there seems to be a mistake when changing the sum from t-1 to t -2
Tariro Manyika
CC BY-SA 4.0
2021-05-11T06:02:56.917Z
Writing out the last term, you have $Q + (1+ r) Q + \ldots + (1+r)^{t-1} Q = Q + (1+ r)[Q + (1+r) Q + \ldots + (1+r)^{t-2} Q] = Q + (1+r) \sum_{j = 0}^{t-2} (1+r) Q$.
tdm
CC BY-SA 4.0
2021-05-12T19:52:46.440Z
Kindly check my answer and tell me what you think , that's how I had done it
Tariro Manyika
CC BY-SA 4.0
2021-05-09T08:41:36.617Z
2021-05-09T08:41:36.617Z
tdm
<p>Let <span class="math-container">$a =$</span> Principle <br />
Let <span class="math-container">$b = 1$</span> + interest rate<br />
Let <span class="math-container">$c_{1} =$</span> future value after year <span class="math-container">$1$</span><br />
Let <span class="math-container">$d =$</span> additional investment</p>
<p><span class="math-container">$c_{1} = ab$</span><br />
<span class="math-container">$c_{2} = (c_{1} + d)b$</span><br />
<span class="math-container">$c_{3} = (c_{2} + d)b$</span><br />
<span class="math-container">$c_{4} = (c_{3} + d)b$</span><br />
<span class="math-container">$c_{5} = (c_{4} + d)b$</span></p>
<p>simplified form for <span class="math-container">$c_{2}$</span></p>
<p><span class="math-container">$c_{2} = (c_{1} + d)b= c_{1}b + db$</span></p>
<p>simplified form for <span class="math-container">$c_{3}$</span></p>
<p><span class="math-container">$c_{3} = (c_{2} + d)b = (c_{1}b + db + d)b = c_{1}b^{2} + db^{2} + db $</span></p>
<p>simplified form for <span class="math-container">$c_{4}$</span></p>
<p><span class="math-container">$c_{4} = (c_{3} + d)b = (c_{1}b^{2} + db^{2} + db + d)b = c_{1}b^{3} + db^{3} + db^{2} + db$</span></p>
<p>simplified form for <span class="math-container">$c_{5}$</span></p>
<p><span class="math-container">$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$</span></p>
<p><span class="math-container">$\therefore$</span></p>
<p>for <span class="math-container">$ n\geq 2 \in \mathbb{N} $</span></p>
<p><span class="math-container">$c_{n} = c_{1}b^{n-1} + db^{n-1} + db^{n-2} + ... + db^{2} + db$</span></p>
<p><span class="math-container">$c_{n} = c_{1}b^{n-1} + d(b^{n-1} + b^{n-2} + ... + b^{2} + b)$</span></p>
<p><span class="math-container">$c_{n} = c_{1}b^{n-1} + d(b + b^{2} + ...+ b^{n-2} + b^{n-1})$</span></p>
<p><span class="math-container">$c_{n} = c_{1}b^{n-1} + d[\sum_{1}^{n-1}b^{x}]$</span></p>
<p><span class="math-container">$c_{n} = (ab)b^{n-1} + d[\sum_{1}^{n-1}b^{x}]$</span></p>
<p><span class="math-container">$c_{n} = ab^{n} + d[\sum_{1}^{n-1}b^{x}]$</span></p>
<p>Using summation of a power series</p>
<p><span class="math-container">$c_{n} = ab^{n} + d[\frac{b(1-b^{n-1})}{1-b}]$</span>, for <span class="math-container">$ n \in \mathbb{N} $</span></p>

CC BY-SA 4.0
2021-05-12T19:51:55.387Z
2021-05-12T21:27:35.530Z
2021-05-12T21:27:35.530Z
Tariro Manyika
<p>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.</p>
<p>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>
<p>P.S. I already solved it using excel but was wondering if there was a simpler formula</p>

<interest-rate>
Is there an equation for compounding interest while adding to the principle amount
Tariro Manyika
# Is there an equation for compounding interest while adding to the principle amount

(Asked by: username_0 on 2021-05-09)

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

**Tags:** interest-rate


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## 2 answers

### Answer by username_1 on 2021-05-09. Score: 4

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:
$$
\begin{align\*}
A\_t &= (1+r)^{t} A\_0 + Q \sum\_{j = 0}^{t-1} (1+ r)^j,\\
&=(1 + r)\left[(1+ r)^{t-1} A\_0 + Q \sum\_{j = 0}^{t-2} (1+ r)^j\right] + Q,\\
&= (1+ r) A\_{t-1} + Q.
\end{align\*}
$$
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:
$$
A\_t = (1+ r)^t A\_0 + Q \frac{(1+ r)^{t} - 1}{r}
$$

> **Comments:**
> **username_0**: Thanks a lot , the way I ended up solving it was a tad pedestrian  
> **username_0**: although I have a problem with this step   \begin{aligned} 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)^{j}\right]+Q, \end{aligned}  
> **username_0**: I see what you are doing but there seems to be a mistake when changing the sum from t-1 to t -2  
> **username_1**: Writing out the last term, you have $Q + (1+ r) Q + \ldots + (1+r)^{t-1} Q = Q + (1+ r)[Q + (1+r) Q + \ldots + (1+r)^{t-2} Q] = Q + (1+r) \sum_{j = 0}^{t-2} (1+r) Q$.  
> **username_0**: Kindly check my answer and tell me what you think , that's how I had done it  

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### Answer by username_0 on 2021-05-12. Score: 1

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 $ n\geq 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} $


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