1. If you invest P 1000 at an annual interest rate of \( 5 \% \) compounded montr how much will you have after 10 years?

To calculate the amount you will have after 10 years with an annual interest rate of 5% compounded monthly, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) is the amount of money accumulated after \( t \) years, - \( P \) is the principal amount (initial investment), - \( r \) is the annual interest rate (in decimal form), - \( n \) is the number of times that interest is compounded per year, - \( t \) is the time the money is invested for in years. Given: - \( P = 1000 \) (initial investment), - \( r = 5\% = 0.05 \) (annual interest rate), - \( n = 12 \) (compounded monthly), - \( t = 10 \) years. Substitute these values into the formula and calculate the amount you will have after 10 years. Calculate the value by following steps: - step0: Calculate: \(1000\left(1+\frac{0.05}{12}\right)^{12\times 10}\) - step1: Divide the terms: \(1000\left(1+\frac{1}{240}\right)^{12\times 10}\) - step2: Add the terms: \(1000\left(\frac{241}{240}\right)^{12\times 10}\) - step3: Multiply the numbers: \(1000\left(\frac{241}{240}\right)^{120}\) - step4: Simplify: \(1000\times \frac{241^{120}}{240^{120}}\) - step5: Rewrite the expression: \(40\times 25\times \frac{241^{120}}{240^{120}}\) - step6: Rewrite the expression: \(40\times 25\times \frac{241^{120}}{40^{120}\times 6^{120}}\) - step7: Reduce the numbers: \(25\times \frac{241^{120}}{40^{119}\times 6^{120}}\) - step8: Rewrite the expression: \(25\times \frac{241^{120}}{5^{119}\times 8^{119}\times 6^{120}}\) - step9: Rewrite the expression: \(5^{2}\times \frac{241^{120}}{5^{119}\times 8^{119}\times 6^{120}}\) - step10: Reduce the numbers: \(\frac{241^{120}}{5^{117}\times 8^{119}\times 6^{120}}\) After 10 years, with an annual interest rate of 5% compounded monthly, you will have approximately $1647.01.