Answer:
The explicit formula that can be used is [tex]A=\$400(1.05)^{2}[/tex]
The account's balance at the beginning of year 3 is [tex]A=\$441[/tex]
Step-by-step explanation:
we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
[tex]t=2\ years\\ P=\$400\\ r=0.05\\n=1[/tex]
substitute in the formula above
[tex]A=\$400(1+\frac{0.05}{1})^{1*2}[/tex]
[tex]A=\$400(1.05)^{2}[/tex]
[tex]A=\$441[/tex]
