Pascals Triangle in exponential? 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?
 A: 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.
A: 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. 
A: 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.
