$1,500 is deposited every year in an account yielding 6% interest compounded annually. How much money will have been saved after 10 years?
The formula for the amount,$A an initial investment of
$P will grow to in t years is given by:
A = P(1+r)^t
The $1500 deposited at the beginning of the 10th year
will have grown to $1500(1+r)^1 in the 1 year it will
have been on deposit at the end of the 10th year.
The $1500 deposited at the beginning of the 9th year
will have grown to $1500(1+r)^2 in the 2 years it will
have been on deposit at the end of the 10th year.
The $1500 deposited at the beginning of the 8th year
will have grown to $1500(1+r)^3 in the 3 years it will
have been on deposit at the end of the 10th year.
...
The $1500 deposited at the beginning of the 2nd year
will have grown to $1500(1+r)^9 in the 9 years it will
have been on deposit at the end of the 10th year.
The $1500 deposited at the beginning of the 1st year
will have grown to $1500(1+r)^10 in the 10 year it will
have been on deposit at the end of the 10th year.
So the sum is this series:
$1500(1+r) + $1500(1+r)^2 + $1500(1+r)^3 + ... + $1500(1+r)^10
and since r = 6% = .06
$1500(1.06) + $1500(1.06)^2 + $1500(1.06)^3 + ... + $1500(1.06)^10
The common ratio is 1.06, so we use the formula for the sum
of the first n terms of a geometric series:
Substituting , and
Edwin