Chapter 3.6, Problem 61E

Solution

a.

To prove: Why after $\text{t}$ years the total amount of interest he receives from his investments will be $f(t)=1000(1+0.06t)$ .

Given information:

The total amount of interest he receives from his investments is $f(t)=1000(1+0.06t)$ .

Proof:

Steve earned an amount of interest after $\text{1}$ year,

  $f(1)=1000+0.06\cdot 1000$

Steve earned an amount of interest after $2$ year,

  $f(2)=1000+0.06\cdot 1000+0.06\cdot 1000$

Steve earned an amount of interest after $n$ year,

  $f(t)=1000+0.06\cdot 1000+0.06\cdot 1000+\cdot \cdot \cdot +0.06\cdot 1000$

Simplifying the above equation to get,

  $\begin{array}{c}f(t)=1000+t(0.06\cdot 1000)\\ =1000+1000\cdot 0.06t\\ f(t)=1000\left(1+0.06t\right)\end{array}$

b.

To compare: The values of the two investments.

Given information:

The total amount of interest he receives from his investments is $f(t)=1000(1+0.06t)$ .

Calculation:

Steve invest compounded annually after $\text{10}$ years is given by,

  $g(t)=1000{\left(1+0.06\right)}^t$

From part (a),

  $f(t)=1000(1+0.06t)$

Now, comparing $g(t)$ , $f(t)$ and tabulate the values for each investment

$t$	$f(t)$	$g(t)$
$\text{1}$	$\$1,060.00$	$\$1,060.00$
$\text{2}$	$\$1,120.00$	$\$1,123.60$
$\text{3}$	$\$1,180.00$	$\$1,191.02$
$\text{4}$	$\$1,240.00$	$\$1,262.48$
$\text{5}$	$\$1,300.00$	$\$1,338.23$
$\text{6}$	$\$1,300.00$	$\$1,418.52$
$\text{7}$	$\$1,420.00$	$\$1,503.63$
$\text{8}$	$\$1,480.00$	$\$1,593.85$
$\text{9}$	$\$1,540.00$	$\$1,689.48$
$\text{10}$	$\$1,600.00$	$\$1,790.85$