Annuities

## **Problem**

An annuity is a series of regular payments that are put into an investment that grows with compounded interest (essentially the sum of compound interest over time). Most books state the annuity formula as

S
n

=

R

[

(

1
i

)

n

−
1

]

i

{\displaystyle S\_n = \frac{R\left[{\left(1 + i \right)}^{n} - 1 \right]}{i} }

.

**Part 1:** A father decides to put $50 at the end of each month in his son's college fund. The interest rate is 3.6% compounded monthly. If his son turns five years old today, how much money will be in the fund when he turns 18 years old? How much more money will be in the fund compared to the situation with no interest rate?

**Part 2:** If the father wanted to save $20000 instead of the result in part 1, how much should he put in at the end of each month?

**Bonus problem**

Begin with the compound interest formula

# S
R
(
1
i

)
n

{\displaystyle S = R(1+i)^n }

. Derive the annuity formula.

## **Solution**

**Part 1**

The payment is $50, so

# R
50

{\displaystyle R = 50 }

.

The interest rate is compounded monthly, so

# i

0.036
12

=
0.003

{\displaystyle i = \frac{0.036}{12} = 0.003 }

.

The number of payments across 13 years (ages 5 - 18) is

# n
13
(
12
# )
156

{\displaystyle n = 13(12) = 156 }

.

Thus,

S

156

=

50

[

(

1
0.003

)

156

−
1

]

0.003

{\displaystyle {S}\_{156} = \frac{50\left[{\left(1 + 0.003 \right)}^{156} - 1 \right]}{0.003} }

S

156

=
9927.98466

{\displaystyle {S}\_{156} = 9927.98466 }

.
The son will have $9927.98 saved up for college by his 18th birthday.

If the "annuity" had no interest,

S

156

=
50
(
156
# )
7800

{\displaystyle {S}\_{156} = 50(156) = 7800 }

.
Subtracting this result to the result with 3.6% interest rate yields

9927.98
−
# 7800
2127.98

{\displaystyle 9927.98 - 7800 = 2127.98 }

.
The interest rate of the annuity contributed an additional $2127.98 to the college fund.

**Part 2**

# 20000

R

[

(

1
0.003

)

156

−
1

]

0.003

{\displaystyle 20000 = \frac{R\left[{\left(1 + 0.003 \right)}^{156} - 1 \right]}{0.003} }

# R

2000
(
0.003
)

[

(

1
0.003

)

156

−
1

]

{\displaystyle R = \frac{2000(0.003)}{\left[{\left(1 + 0.003 \right)}^{156} - 1 \right]} }

# R

60
0.595679

{\displaystyle R = \frac{60}{0.595679} }

# R
100.7253907

{\displaystyle R = 100.7253907 }

The father should put in $100.73 at the end of each month to reach $20000 by the time his son turns 18.

**Bonus problem**

On the first payment

S
0

=
R

{\displaystyle S\_0 = R }

.

On the second payment

S
1

=
R

{\displaystyle S\_1 = R }

, and the first payment grew due to interest

S
0

=
R
(
1
i
)

{\displaystyle S\_0 = R(1+i) }

.

This recursive pattern continues for

n

{\displaystyle n}

 payments

[
0
,
n
−
1
]

{\displaystyle [0,n-1] }

,

S
n

=

S
0

S
1

S
3

.
.
.

S

n
−
1

{\displaystyle S\_n = S\_0 + S\_1 + S\_3 + ... + {S}\_{n-1} }

S
n

=
R

(
1
i
)

n
−
1

.
.
.
R
(
1
i
)
R

{\displaystyle S\_n = R{(1+i)}^{n-1} + ... + R(1+i) + R }

.
Using the geometric series formula

∑

# k
0

n
−
1

a

r
k

=

a
(

r
n

−
1
)

r
−
1

{\displaystyle \sum\_{k=0}^{n-1} ar^k = \frac{a(r^n - 1 )}{r-1} }

,
where

# a
R

{\displaystyle a = R }

 and

# r
(
1
i
)

{\displaystyle r = (1+i) }

S
n

=

R
(
(
1
i

)
n

−
1
)

(
1
i
)
−
1

{\displaystyle S\_n = \frac{R((1+i)^n - 1 )}{(1+i)-1} }

.
Therefore,

S
n

=

R
(
(
1
i

)
n

−
1
)

i

{\displaystyle S\_n = \frac{R((1+i)^n - 1 )}{i} }

.