Find the sum of the finite geometric series. mr. jamison deposited $100 into a new savings account on january 1. on the first day of each month thereafter, he deposited three times the amount he deposited in the previous month. on june 15 of the same year, the total amount mr. jamison has deposited is $ .

Work Shown:

Triple each deposit amount as you go from month to month.

100 deposited in January

300 deposited in February

900 deposited in March

2700 deposited in April

8100 deposted in May

24300 deposited in June

Add up those amounts: 100+300+900+2700+8100+24300 = 36,400

A shortcut is to use the following formula

S = a*(1-r^n)/(1-r)

where

S = sum of the first n terms of a geometric sequence

a = first term = 100 which is the amount deposited in Jan.

r = common ratio = 3 is the idea we triple each deposit

n = number of terms = 6 is the number of months from Jan to June

So we get,

S = a*(1-r^n)/(1-r)

S = 100*(1-3^6)/(1-3)

S = 36,400

which is the same answer as before. This method is much faster if n is large.

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

$6.42

step-by-step explanation: