Question 1005042
month 1 is 10 dollars.
month 2 is 2 * 10 = 20 dollars.
month 2 is 2 * 20 = 40 dollars.
month 4 is 2 * 40 = 80 dollars.


if you look at what's happening to your original deposit, you get:


month 1 = 10 * 2^0 = 10 * 1 = 10
month 2 = 10 * 2^1 = 10 * 2 = 20
month 3 = 10 * 2^2 = 10 * 4 = 40
month 4 = 10 * 2^3 = 10 * 8 = 80
month 5 = 10 * 2^4 = 10 * 16 = 160


month n would therefore be equal to 10 * 2^(n-1)


this looks very much like a geometric series where A1 is the initial amount, and An is the amount in time point n.


the general formula for a geometric series is An = A1 * r^(n-1)


An is the amount in time point n.
A1 is the original amount in time point 1.
r is the common ratio
n is the time point.


using this formula, you get:


A1 = 10 * 2^(1-1) = 10 * 2^0 = 10


A2 = 10 * 2^(2-1) = 10 * 2^1 = 10 * 2 = 20


A3 = 10 * 2^(3-1) = 10 * 2^2 = 10 * 4 = 40


A4 = 10 * 2^(4-1) = 10 * 2^3 = 10 * 8 = 80


A5 = 10 * 2^(5-1) = 10 * 2^4 = 10 * 16 = 160


money is doubling every month.


looks like the formula is what you're looking for.


in functional notation, you would say:


f(n) = 10 * 2^(n-1)


your function name is f.
the argument to your function is n.
your independent variable is n.
your dependent variable is f(n).


give it a different function name, such as g, and your function becomes:


g(n) = 10 * 2^(n-1)


your function name is g.
the argument to your function is n.
your independent variable is n.
your dependent variable is g(n).



give it a different argument name, such as x, and your function becomes:


g(x) = 10 * 2^(x-1)


your function name is g.
the argument to your function is x.
your independent variable is x.
your dependent variable is g(x).



f(n) = 10 * 2^(n-1) will give you the same result as g(x) = 10 * 2^(x-1) as long as x = n.