1. Stated Problem: Tumi deposits 500 every quarter for 12 years in an account with an 8% annual interest rate compounded quarterly. We want to find the account balance right after the last deposit. 2. Identify variables: - Quarterly deposit $P = 500$ - Annual interest rate $r = 0.08$ - Quarterly interest rate $i = r / 4 = 0.08 / 4 = 0.02$ - Number of quarters $n = 12 \times 4 = 48$ 3. Since deposits happen at the end of each quarter and interest compounds quarterly, the future value of an ordinary annuity formula applies: $$ FV = P \times \frac{(1+i)^n - 1}{i} $$ 4. Substitute the known values: $$ FV = 500 \times \frac{(1+0.02)^{48} - 1}{0.02} $$ 5. Calculate $(1+0.02)^{48}$: $$ (1.02)^{48} \approx 2.5937424601 $$ 6. Compute numerator: $$ 2.5937424601 - 1 = 1.5937424601 $$ 7. Divide by $i$: $$ \frac{1.5937424601}{0.02} = 79.687123005 $$ 8. Multiply by deposit $P$: $$ 500 \times 79.687123005 = 39843.5615 $$ 9. Round to two decimals: The final amount in the account immediately after the last deposit is approximately $39843.56$.