Thread: finance question plz need help

Paul deposits $6,000 at the end of each year for 30 years in an account paying 6% interest compounded annually.
i) Find the final amount he will have on deposit.

ii) 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?
i
ii) How much would Paul lose over 30 years by using his sister’s bank?

Originally Posted by gracy

Paul deposits $6,000 at the end of each year for 30 years in an account paying 6% interest compounded annually.
i) Find the final amount he will have on deposit.

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

ii) 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?

In this case the 1.06 is replaced by 1.0585

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

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

ii) How much would Paul lose over 30 years by using his sister’s bank?

$474349.12 - $465010.39 = $9338.73

RonL

(check the arithmetic and algebra if possible)