Pascals Triangle in exponential?

Question

Here is a problem:

suppose you have 40 gallons of red-dyed water and you add 1 gallon of blue-dyed water then take out a gallon from the mixture. Supposing that the two waters completely mixed before taking out a gallon, how many times must one take out a gallon of red-dyed and add a gallon of blue-dyed to have 20 gallons of each?

Before seeing the exponential decrease, I figured like this:

after first time: 40 - 40/41

after second time: (40 - 40/41) - (40 - 40/41)/41 = 40 - 2(40/41) + 40/41^2

after third time: (40 - 40/41) - (40 - 40/41)/41 - (40 - 40/41) - ((40 - 40/41)/41)/41 = 40 - 3(40/41) + 3(40/41^2) - 40/(41^3)

As one can see the coefficients of each expression matches Pascals Triangle, Why? Also, why is it adding then subtracting each term?

Answer 1

Each time, you're multiplying the old proportion of red-dyed water by $\frac{40}{41}$. If there's $a_n$ litres of red dye after $n$ steps, then you get $\frac{40}{41}a_n$ gallons of red dye and $\frac{40}{41}(40-a_n)+\frac{1}{41}$ gallons of blue dye. So, ignoring the blue dye, your red dye after $n$ steps is $40\left(1-\frac{40}{41}\right)^n$ gallons.

Answer 2

You obtain each term from the previous by subtracting 1/41 the previous value. Thus you are multiplying by (1-1/41) each time. Thus the n'th term will be

40$(1-1/41)^n$.

In general: $$ (a-b)^n= {n \choose 0}a^n-{n \choose 1}a^{n-1}b+{n \choose 2}a^{n-2}b^2\cdots +(-1)^n {n \choose n}b^n $$

Here the numbers ${n\choose r}$ are the values on the $n$th layer of Pascal's triangle.

This is an instance of the binomial theorem.

Answer 3

Your left hand side can be written as: $$(1-x)$$ $$(1-x)-x(1-x)$$ $$(1-x)(1-x)-x(1-x)(1-x)$$ where $x=\frac1{41}$, and the factor of $40$ is removed.

This can be simplified to:

$$(1-x)$$ $$\color{red}{(1-x)}-x\color{red}{(1-x)}=(1-x)\color{red}{(1-x)}=(1-x)^2$$ $$\color{red}{(1-x)(1-x)}-x\color{red}{(1-x)(1-x)}=(1-x)\color{red}{(1-x)(1-x)}=(1-x)^3$$

by applying distributivity over the red parts.