Let's assume that we deposit
P dollars into an account. When the interest is compounded yearly with the interest rate equal to
r%, the balance after
t years can be modeled with the formula for .
y=P(1+r)t
In the first year we deposit
P=2000 dollars in an account with interest rate
r=5%=0.05. Let
y30 be the amount of money after
t=30 years from that deposit. Let's find
y30.
y30=P(1+r)t
y30=2000(1+0.05)30
y30=2000(1.05)30
Every year for
30 years we deposit an additional
$2000 into the account. Let's look at how much money we will get from the deposit after the first year. We also deposit
P=2000 dollars with the same interest rate
r=0.05. Let
y29 be the amount of money after
t=29 years. Let's find
y29.
y29=P(1+r)t
y29=2000(1+0.05)29
y29=2000(1.05)29
Therefore the total amount of our money after
30 years is equal to the following partial sum of geometric series.
2000(1.05)30+2000(1.05)29+…+2000(1.05)1+2000(1.05)0
Let's rearrange terms of the series.
2000(1.05)0+2000(1.05)1+…+2000(1.05)29+2000(1.05)30
We will use the formula for finding .
Sn=1−Ra1(1−Rn)
For this series, the first term is
a1=2000 and the is
R=1.05. Since we are adding
31 terms to represent the total
31 years (including the first year you deposited),
n=31. Let's substitute the values and find the total amount of money in our account after
30 years.
Sn=1−Ra1(1−Rn)
S31=1−1.052000(1−(1.05)31)
S31=-0.052000(1−(1.05)31)
S31≈141521.58
Therefore, after
30 years we will have about
$141521.58 in our account.