If you invest $\$10$ at beginning of each month for $10$ years, how much will you have at end of the $10$ years?

Question

Problem

If you invest a dollar at “$6\%$ interest compounded monthly,” it amounts to $(1.005)^n$ dollars after $n$ months. If you invest $\$10$ at the beginning of each month for $10$ years ($120$ months), how much will you have at the end of the $10$ years?

[Source : Mary Boas Mathematical Method in the Physical Sciences $3$Ed: $1.1.13$]

My Understanding

It's a monthly compound interest problem where you start with $\$10$ with a interest rate of $6\%$ and every month you invest an additional $\$10$. The question asks to find the final amount after $10$ years of investing ($120$ months).

I am confused after looking at the solution manual

It takes the sum of the series $$S=10(1.005)+10(1.005)^2+10(1.005)^3+...10(1.005)^{120}$$ and uses the partial sum formula for the geometric series to calculate the sum as $1646.99$

When I was solving the problem, my series wasn't geometric: $$S=10(1.005) + (10(1.005)+10)1.005 + ((10(1.005)+10)1.005)1.005+...$$ which simplifies to $$10(1.005+(1.005^2+1.005)+(1.005^3+1.005^2+1.005)+...)$$

I have two questions.

1. Why it is the case the solution represents this problem with a geometric series if $a$, the initial amount, alters due to the additional $\$10$ investment each month.
2. Is there a formula to calculating the sum of the series I wrote?

Thanks!

Answer 1

For part (a), I think that you are looking at it from the wrong perspective. Suppose that instead of making monthly deposits into one account, each monthly deposit went into a separate bank account. So, you end up with $(120)$ separate bank accounts. Then, the issue is, what is the total value of all of these accounts, after $(10)$ years.

If an account is opened with $(k)$ investment periods remaining, with $k \in \{1,2,\cdots, 120\}$, then after 10 years, that specific account will have grown to $10(1.005)^k$. This explains the offered solution.

As far as your 2nd question:

• It is difficult to decipher, because you didn't use mathJax to format the math.

• I am unable to understand what analysis you used to conjure the math expression that corresponds to your 2nd question.

• If, after reading this answer, you are able to edit your original posting with MathJax, and you then (still) want to try to find a nice closed form expression that represents the math formula re your 2nd question, please leave a comment following my answer. Then, I will take a crack at it.

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Addendum
A more formal (i.e. less intuitive) attack on part (a) may be done via induction.

After $(1)$ month, your total balance is

$\displaystyle (10)[(1.005)^1 + 1].$

Suppose that after $(K)$ months, your total balance is

$\displaystyle(10)\left[\sum_{i=0}^{K} (1.005)^i\right].$

Then, in month $(K+1)$ your balance will become

$\displaystyle (10) \times \left\{\left[(1.005) \times \sum_{i=0}^{K} (1.005)^i\right] + 1\right\} = (10) \times \left[\sum_{i=0}^{K+1} (1.005)^i\right] .$

Answer 2

Your series $$ S = 10(1.005+(1.005^2+1.005)+(1.005^3+1.005^2+1.005)+...+ (1.005^k + 1.005^{k-1} + \ldots + 1.005)), $$ can be simplified a good deal. I'm going to use $r$ instead of $1.005$, so that \begin{align} S & = 10\bigl(r+(r^2+r)+(r^3+r^2+r)+...+ (r^k + r^{k-1} + \ldots + r)\bigr)\\ & = 10r\bigl(1+(r+1)+(r^2+r+1)+...+ (r^{k-1} + \ldots + 1)\bigr)\\ \end{align} Each inside terms is a geometric series (in reverse order, but no big deal), so we get \begin{align} S & = 10r\bigl(1+(r+1)+(r^2+r+1)+...+ (r^{k-1} + \ldots + 1)\bigr)\\ & = 10r\bigl(\frac{1-r}{1-r}+\frac{1-r^2}{1-r}+\frac{1-r^3}{1-r}+...+ \frac{1-r^k}{1-r}\bigr)\\ & = \frac{10r}{1-r}\bigl(1-r+1-r^2+1-r^3+...+ 1-r^k\bigr)\\ & = \frac{10r}{1-r}\bigl(1+1_ \ldots + 1\bigr) + \frac{10r}{1-r}\bigl(-r-r^2-r^3 -\ldots -r^k\bigr)\\ & = \frac{10r}{1-r}k - \frac{10r^2}{1-r}\bigl(1+r+ \ldots + r^{k-1}\bigr)\\ & = \frac{10kr}{1-r} - \frac{10r^2}{1-r}\frac{1-r^k}{1-r}\\ & = \frac{10kr}{1-r} - \frac{10r^2}{(1-r)^2}(1-r^k). \end{align} I'm sure that can be simplified a little further, but the gist is "be careful, and use the sum-of-a-geometric-series formula over and over."