# 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 &= (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: $$A_t = (1+ r)^t A_0 + Q \frac{(1+ r)^{t} - 1}{r}$$

• Thanks a lot , the way I ended up solving it was a tad pedestrian May 10 at 9:25
• 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} May 10 at 9:43
• I see what you are doing but there seems to be a mistake when changing the sum from t-1 to t -2 May 10 at 9:44
• 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
May 11 at 6:02
• Kindly check my answer and tell me what you think , that's how I had done it May 12 at 19:52

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 $$n \in \mathbb{N}$$