Find the final amount in the following retirement account, in which the rate of return on the account and the regular contribution change over time. \( \$ 1400 \) per quarter invested at \( 5.1 \% \), compounded quarterly, for 12 years; then \( \$ 1600 \) per quarter invested at \( 6.3 \% \), compounded quarterly, for 15 years. Find the final amount in the account. \( \$ \square \) (Do not round until the final answer. Then round to the nearest dollar as needed.)

To find the final amount in the retirement account, we will break the problem into two parts based on the different investment periods and rates of return. ### Part 1: First Investment Period 1. **Initial Contribution**: \( P_1 = 1400 \) (quarterly) 2. **Rate of Return**: \( r_1 = 5.1\% = 0.051 \) (annual) 3. **Compounding Frequency**: \( n = 4 \) (quarterly) 4. **Time Period**: \( t_1 = 12 \) years The formula for the future value of a series of cash flows (ordinary annuity) is given by: \[ FV_1 = P \times \frac{(1 + r/n)^{nt} - 1}{r/n} \] Where: - \( P \) is the payment amount per period - \( r \) is the annual interest rate - \( n \) is the number of compounding periods per year - \( t \) is the number of years Substituting the values for the first investment period: \[ FV_1 = 1400 \times \frac{(1 + 0.051/4)^{4 \times 12} - 1}{0.051/4} \] ### Part 2: Second Investment Period 1. **Initial Contribution**: \( P_2 = 1600 \) (quarterly) 2. **Rate of Return**: \( r_2 = 6.3\% = 0.063 \) (annual) 3. **Compounding Frequency**: \( n = 4 \) (quarterly) 4. **Time Period**: \( t_2 = 15 \) years The future value for the second investment period will be calculated similarly, but we need to consider the future value from the first period as the initial amount for the second period. The future value for the second investment period is given by: \[ FV_2 = FV_1 \times (1 + r_2/n)^{nt_2} + P_2 \times \frac{(1 + r_2/n)^{nt_2} - 1}{r_2/n} \] Now, we will calculate both future values \( FV_1 \) and \( FV_2 \). Let's perform the calculations. Calculate the value by following steps: - step0: Calculate: \(\left(1+\frac{0.063}{4}\right)^{4\times 15}\) - step1: Divide the terms: \(\left(1+\frac{63}{4000}\right)^{4\times 15}\) - step2: Add the numbers: \(\left(\frac{4063}{4000}\right)^{4\times 15}\) - step3: Multiply the numbers: \(\left(\frac{4063}{4000}\right)^{60}\) - step4: Use the properties of exponents: \(\frac{4063^{60}}{4000^{60}}\) Calculate or simplify the expression \( 1400 * ((1 + 0.051/4)^(4*12) - 1) / (0.051/4) \). Calculate the value by following steps: - step0: Calculate: \(\frac{1400\left(\left(1+\frac{0.051}{4}\right)^{4\times 12}-1\right)}{\left(\frac{0.051}{4}\right)}\) - step1: Remove the parentheses: \(\frac{1400\left(\left(1+\frac{0.051}{4}\right)^{4\times 12}-1\right)}{\frac{0.051}{4}}\) - step2: Divide the terms: \(\frac{1400\left(\left(1+\frac{51}{4000}\right)^{4\times 12}-1\right)}{\frac{0.051}{4}}\) - step3: Add the numbers: \(\frac{1400\left(\left(\frac{4051}{4000}\right)^{4\times 12}-1\right)}{\frac{0.051}{4}}\) - step4: Multiply the numbers: \(\frac{1400\left(\left(\frac{4051}{4000}\right)^{48}-1\right)}{\frac{0.051}{4}}\) - step5: Subtract the numbers: \(\frac{1400\times \frac{4051^{48}-4000^{48}}{4000^{48}}}{\frac{0.051}{4}}\) - step6: Divide the terms: \(\frac{1400\times \frac{4051^{48}-4000^{48}}{4000^{48}}}{\frac{51}{4000}}\) - step7: Multiply the numbers: \(\frac{\frac{7\times 4051^{48}-7\times 4000^{48}}{200^{47}\times 20^{48}}}{\frac{51}{4000}}\) - step8: Multiply by the reciprocal: \(\frac{7\times 4051^{48}-7\times 4000^{48}}{200^{47}\times 20^{48}}\times \frac{4000}{51}\) - step9: Rewrite the expression: \(\frac{7\times 4051^{48}-7\times 4000^{48}}{200^{47}\times 20^{48}}\times \frac{200\times 20}{51}\) - step10: Reduce the numbers: \(\frac{7\times 4051^{48}-7\times 4000^{48}}{200^{46}\times 20^{48}}\times \frac{20}{51}\) - step11: Rewrite the expression: \(\frac{7\times 4051^{48}-7\times 4000^{48}}{20^{46}\times 10^{46}\times 20^{48}}\times \frac{20}{51}\) - step12: Reduce the numbers: \(\frac{7\times 4051^{48}-7\times 4000^{48}}{20^{45}\times 10^{46}\times 20^{48}}\times \frac{1}{51}\) - step13: Multiply the fractions: \(\frac{7\times 4051^{48}-7\times 4000^{48}}{20^{93}\times 10^{46}\times 51}\) - step14: Multiply: \(\frac{7\times 4051^{48}-7\times 4000^{48}}{51\times 20^{93}\times 10^{46}}\) Calculate or simplify the expression \( 91905.386672 * 2.553935 \). Calculate the value by following steps: - step0: Calculate: \(91905.386672\times 2.553935\) - step1: Convert the expressions: \(\frac{5744086667}{62500}\times \frac{510787}{200000}\) - step2: Multiply the fractions: \(\frac{5744086667\times 510787}{62500\times 200000}\) - step3: Multiply the terms: \(\frac{2934004796376929}{62500\times 200000}\) - step4: Multiply the terms: \(\frac{2934004796376929}{12500000000}\) Calculate or simplify the expression \( 1600 * ((1 + 0.063/4)^(4*15) - 1) / (0.063/4) \). Calculate the value by following steps: - step0: Calculate: \(\frac{1600\left(\left(1+\frac{0.063}{4}\right)^{4\times 15}-1\right)}{\left(\frac{0.063}{4}\right)}\) - step1: Remove the parentheses: \(\frac{1600\left(\left(1+\frac{0.063}{4}\right)^{4\times 15}-1\right)}{\frac{0.063}{4}}\) - step2: Divide the terms: \(\frac{1600\left(\left(1+\frac{63}{4000}\right)^{4\times 15}-1\right)}{\frac{0.063}{4}}\) - step3: Add the numbers: \(\frac{1600\left(\left(\frac{4063}{4000}\right)^{4\times 15}-1\right)}{\frac{0.063}{4}}\) - step4: Multiply the numbers: \(\frac{1600\left(\left(\frac{4063}{4000}\right)^{60}-1\right)}{\frac{0.063}{4}}\) - step5: Subtract the numbers: \(\frac{1600\times \frac{4063^{60}-4000^{60}}{4000^{60}}}{\frac{0.063}{4}}\) - step6: Divide the terms: \(\frac{1600\times \frac{4063^{60}-4000^{60}}{4000^{60}}}{\frac{63}{4000}}\) - step7: Multiply the numbers: \(\frac{\frac{4063^{60}-4000^{60}}{2^{294}\times 5^{178}}}{\frac{63}{4000}}\) - step8: Multiply by the reciprocal: \(\frac{4063^{60}-4000^{60}}{2^{294}\times 5^{178}}\times \frac{4000}{63}\) - step9: Rewrite the expression: \(\frac{4063^{60}-4000^{60}}{2^{294}\times 5^{178}}\times \frac{32\times 125}{63}\) - step10: Rewrite the expression: \(\frac{4063^{60}-4000^{60}}{2^{294}\times 5^{178}}\times \frac{2^{5}\times 125}{63}\) - step11: Reduce the numbers: \(\frac{4063^{60}-4000^{60}}{2^{289}\times 5^{178}}\times \frac{125}{63}\) - step12: Rewrite the expression: \(\frac{4063^{60}-4000^{60}}{2^{289}\times 5^{178}}\times \frac{5^{3}}{63}\) - step13: Reduce the numbers: \(\frac{4063^{60}-4000^{60}}{2^{289}\times 5^{175}}\times \frac{1}{63}\) - step14: Multiply the fractions: \(\frac{4063^{60}-4000^{60}}{2^{289}\times 5^{175}\times 63}\) - step15: Multiply: \(\frac{4063^{60}-4000^{60}}{63\times 2^{289}\times 5^{175}}\) Calculate or simplify the expression \( 234720.38371 + 157860.07546 \). Calculate the value by following steps: - step0: Calculate: \(234720.38371+157860.07546\) - step1: Add the numbers: \(392580.45917\) Now, let's summarize the calculations step by step: ### Part 1: First Investment Period 1. **Future Value Calculation**: \[ FV_1 = 1400 \times \frac{(1 + 0.051/4)^{4 \times 12} - 1}{0.051/4} \approx 91905.39 \] ### Part 2: Second Investment Period 1. **Future Value of the First Investment After 15 Years**: \[ FV_1 \text{ grows to } FV_1 \times (1 + 0.063/4)^{4 \times 15} \approx 91905.39 \times 2.553935 \approx 234720.38 \] 2. **Future Value Calculation for the Second Investment**: \[ FV_2 = 1600 \times \frac{(1 + 0.063/4)^{4 \times 15} - 1}{0.063/4} \approx 157860.08 \] ### Final Amount Calculation Now, we add the future values from both investment periods: \[ \text{Total Final Amount} = FV_1 \text{ (after 15 years)} + FV_2 \approx 234720.38 + 157860.08 \approx 392580.46 \] ### Final Answer Rounding to the nearest dollar, the final amount in the retirement account is: \[ \boxed{392580} \]