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Answer :
Barry makes three transactions each month on his checking account:
1. He deposits \[tex]$700.
2. He withdraws \$[/tex]150 for gas.
3. He withdraws \[tex]$400 for other expenses.
To find the net change in his balance for each month, we compute:
$[/tex][tex]$
700 - 150 - 400 = 150.
$[/tex][tex]$
This means that every month his balance increases by \$[/tex]150. Given that his account balance is \[tex]$1900 at the end of the first month, we can model this situation with the following recursive formula:
$[/tex][tex]$
f(1)=1900
$[/tex][tex]$
$[/tex][tex]$
f(n)=f(n-1)+150\quad \text{for } n\geq 2.
$[/tex][tex]$
This recursive equation captures that each month the balance is increased by \$[/tex]150 from the previous month’s balance.
Therefore, the correct option is the one that states:
[tex]$$
f(1)=1900,\quad f(n)=f(n-1)+150\quad \text{for } n \geq 2.
$$[/tex]
This corresponds to Option C.
1. He deposits \[tex]$700.
2. He withdraws \$[/tex]150 for gas.
3. He withdraws \[tex]$400 for other expenses.
To find the net change in his balance for each month, we compute:
$[/tex][tex]$
700 - 150 - 400 = 150.
$[/tex][tex]$
This means that every month his balance increases by \$[/tex]150. Given that his account balance is \[tex]$1900 at the end of the first month, we can model this situation with the following recursive formula:
$[/tex][tex]$
f(1)=1900
$[/tex][tex]$
$[/tex][tex]$
f(n)=f(n-1)+150\quad \text{for } n\geq 2.
$[/tex][tex]$
This recursive equation captures that each month the balance is increased by \$[/tex]150 from the previous month’s balance.
Therefore, the correct option is the one that states:
[tex]$$
f(1)=1900,\quad f(n)=f(n-1)+150\quad \text{for } n \geq 2.
$$[/tex]
This corresponds to Option C.
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