Select the correct answer.

Each month, Barry makes three transactions in his checking account:
- He deposits [tex]$\$700$[/tex] from his paycheck.
- He withdraws [tex]$\$150$[/tex] to buy gas for his car.
- He withdraws [tex]$\$400$[/tex] for other expenses.

If his account balance is [tex]$\$1,900$[/tex] at the end of the 1st month, which recursive equation models Barry's account balance at the end of month [tex]$m$[/tex]?

A. [tex]f(1) = 1,900[/tex]
[tex]f(n) = 150 \cdot f(n-1)[/tex], for [tex]n \geq 2[/tex]

B. [tex]f(1) = 1,900[/tex]
[tex]f(n) = f(n-1) + 150[/tex], for [tex]n \geq 2[/tex]

C. [tex]f(1) = 1,900[/tex]
[tex]f(n) = f(n-1) + 700[/tex], for [tex]n \geq 2[/tex]

D. [tex]f(1) = 1,900[/tex]
[tex]f(n) = f(n-1) - 150[/tex], for [tex]n \geq 2[/tex]

Certainly! Let's work through the transactions to determine the recursive equation for Barry's account balance at the end of each month.

Barry performs these transactions each month:
1. Deposits [tex]$700.
2. Withdraws $[/tex]150 for gas.
3. Withdraws [tex]$400 for other expenses.

First, we need to figure out the net change to his account balance each month:

- The total withdrawal each month is $[/tex]150 + [tex]$400 = $[/tex]550.
- Since Barry deposits [tex]$700 every month, we calculate the net change by subtracting the total withdrawals from the deposit: $[/tex]700 - [tex]$550 = $[/tex]150.

This means each month, after all transactions, Barry's account balance increases by [tex]$150.

Based on this, we can set up the recursive formula:
- The initial balance at the end of the first month is $[/tex]1,900. So, we start with [tex]$ f(1) = 1,900 $[/tex].
- Each subsequent month, his balance increases by $150, which gives us the recursive formula: [tex]$ f(n) = f(n-1) + 150 $[/tex] for [tex]$ n \geq 2 $[/tex].

Therefore, the correct choice is answer B:
[tex]\[ f(1) = 1,900 \][/tex]
[tex]\[ f(n) = f(n-1) + 150, \text{ for } n \geq 2 \][/tex]