note that this isn't a summation problem, it just wants to find on what day will she get 10,000,000 dollars in 1 day
convert that the pennies to make it easier (100 pennies=1 dollar so that's 1,000,000,000 or a billion pennies) (note that 1,000,000,000=10^9)
each day is 3 times of the other
first day is 1, 2nd is 1*3, 3rd day is 1*3*3, 4th day is 1*3*3*3, etc
hmm, so after n days, the number of pennies is

one long way is to keep dividing by 3 to find how many times we need to do that to find n
easier way is to do this:
1,000,000,000=

take ln of both sides


use properties of logarithms

divide both sides by ln(3)

add 1 to both sides

using your calculator we get 19.863=n
so after 19.863 days
but we need a whole number of days so round up to 20
20 days
it will be on day 20 of the month