Math is not useless, it is just you don't know how to use it

One day, my father just wanted to make sure that I know my financial formulas and calculations, so he proposed a scenario :

"assume that someone has deposited a 100 bucks into a bank, and the interest rate for every year is 5%, how much money will he have after 30 yrs ? assume someone else dump in 100 buck at the beginning, same bank, same interest rate, except that he dump in 50 buck for every year, how much money will he have after 30 yrs ?"

Before diving into the solutions, let's name a few things,

d = Deposit, here it is 100 bucks

I = interest rate, in this case, it would be 1.05

s = savings, in this case, 50 bucks

T = total money

; for indexing

---

Question 1

The first part is extremely easy and straightforward, we can go

T;0 = dI

T;1 = dII

T;2 = dIII

...

T;n = d × I^n

So here, n = 30

T;30 = 100 × 1.05^30 ≈ 432.19

---

Question 2

This one is a little bit challenging, naively, we might have the following ...

T;n = (...(((d)I + s)I + s)I ... s)I

Although this can be treated recursively, it is not easy to calculate by using the recursive mindset. Here, we might have to do a little bit of brute forcing and pattern seeking ...

here we might have 2 versions : either the interest rate works at the beginning of the year or at the end of the year.

If the interest rate works at the beginning of the year :

T;0 = d

T;1 = dI + s

T;2 = dI^2 + sI + s

T;3 = dI^3 + sI^2 + sI + s

T;4 = dI^4 + sI^3 + sI^2 + sI + s

...

T;n = dI^n + sI^(n-1) + sI^(n-2) + sI(n-3) + ... + sI^3 + sI^2 + sI^1 + s

T;n = dI^n + s[ I^(n-1) + I^(n-2) + I^(n-3) + ... + I^3 + I^2 + I^1 + I^0 ]

T;n = dI^n + s[ I^0 + I^1 + I^2 + ... + I^(n-1) ]

If the interest rate works at the end of the year :

T;0 = dI + sI

T;1 = dI^2 + sI^2 + sI

T;2 = dI^3 + sI^3 + sI^2 + sI

T;3 = dI^4 + sI^4 + sI^3 + sI^2 + sI

T;4 = dI^5 + sI^5 + sI^4 + sI^3 + sI^2 + sI

...

T;n = dI^(n+1) + sI^(n+1) + sI^n + sI^(n-1) + ... + sI^3 + sI^2 + sI^1

T;n = dI^(n+1) + s[ I^(n+1) + I^n + I^(n-1) + ... + I^3 + I^2 + I^1 ]

T;n = dI^(n+1) + s[ I^1 + I^2 + I^3 + ... + I^(n+1) ]

We can see both scenarios are quite similar, the only difference is the red part.

And it is been a while since I saw this red part, this sequence seems quite popular, when I Googled it, Google said that this is a Geometric Sequence. And I rushed to my Junior3 Math Textbook, and here is the proof for getting the sum of a geometric sequence :

let S = a + ar + ar^2 + ar^3 + ... + ar^n

∴ Sr = ar + ar^2 + ar^3 + ar^4 + ... + ar^(n+1)

∴ Sr - S = ar^(n+1) - ar

S(r-1) = a(r^(n+1) - r)

S = a × (r^(n+1) - r)/(r-1)

S = ar × (r^n - 1) / (r - 1)

So by using this, both formulas become:

T;beginning;n

= dI^n + s[ I^0 + I^1 + I^2 + ... + I^(n-1) ]

= dI^n + s[ 1I × (I^(n-1)-1)/(I-1) ]

T;endding;n

= dI^(n+1) + s[ I^1 + I^2 + I^3 + ... + I^(n+1) ]

= dI^(n+1) + s[ I^2 × (I^n-1)/(I-1) ]

Now we can plug in the numbers, I will not do it here thou.

---

Back when I was 15, none of us knew how to use the geometric sequence.

And fast forward now, I have this thing on my bookshelf.

Another proof for me to keep my high school STEM textbooks.