Consider the $6000 deposited at the end of year 1, its worth $6000

at the end of year 1, $6000*1.06 at the end of year 2, and $6000*(1.06)^(n-1)

at the end of the n-th year.

So the first years money is worth $6000*(1.06)^29 at the end of the 30-th

year. the second years money is worth $6000*(1.06)^28 at the end of the

30-th year, and so on.

Therefore the total in the account at the end of 30 years is:

$6000*[ (1.06)^29 + (1.06)^28 + ... + (1.06)^2 + 1.06 +1]

................. = $6000 [1-1.06^30]/[1-1.06] = $474349.12

In this case the 1.06 is replaced by 1.0585ii) Paul’s sister works in a bank that pays 5.85% compounded annually. If he

deposits his money in this bank instead of the one above, how much will he

have in his account?

$6000*[ (1.0585)^29 + (1.0585)^28 + ... + (1.0585)^2 + 1.0585 +1]

................. = $6000 [1-1.0585^30]/[1-1.0585] = $465010.39

$474349.12 - $465010.39 = $9338.73ii) How much would Paul lose over 30 years by using his sister’s bank?

RonL

(check the arithmetic and algebra if possible)