1. **Problem statement:** Calculate the interest earned on 20000 invested for 4 years at 4% interest compounded in different ways. 2. **Formula for compound interest:** For compounding $n$ times per year, the amount $A$ is given by: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where $P$ is the principal, $r$ is the annual interest rate (decimal), $n$ is the number of compounding periods per year, and $t$ is the time in years. 3. **Formula for continuous compounding:** $$A = Pe^{rt}$$ 4. **Given:** - $P = 20000$ - $r = 0.04$ - $t = 4$ --- **a. Annually ($n=1$):** $$A = 20000 \left(1 + \frac{0.04}{1}\right)^{1 \times 4} = 20000 (1.04)^4$$ Calculate: $$20000 (1.04)^4 = 20000 \times 1.16985856 = 23397.17$$ Interest earned: $$23397.17 - 20000 = 3397.17$$ **b. Semiannually ($n=2$):** $$A = 20000 \left(1 + \frac{0.04}{2}\right)^{2 \times 4} = 20000 (1.02)^8$$ Calculate: $$20000 (1.02)^8 = 20000 \times 1.171659 = 23433.18$$ Interest earned: $$23433.18 - 20000 = 3433.18$$ **c. Quarterly ($n=4$):** $$A = 20000 \left(1 + \frac{0.04}{4}\right)^{4 \times 4} = 20000 (1.01)^{16}$$ Calculate: $$20000 (1.01)^{16} = 20000 \times 1.172578 = 23451.56$$ Interest earned: $$23451.56 - 20000 = 3451.56$$ **d. Monthly ($n=12$):** $$A = 20000 \left(1 + \frac{0.04}{12}\right)^{12 \times 4} = 20000 (1.0033333)^{48}$$ Calculate: $$20000 (1.0033333)^{48} = 20000 \times 1.172974 = 23459.48$$ Interest earned: $$23459.48 - 20000 = 3459.48$$ **e. Continuously compounded:** $$A = 20000 e^{0.04 \times 4} = 20000 e^{0.16}$$ Calculate: $$20000 \times 1.17351 = 23470.20$$ Interest earned: $$23470.20 - 20000 = 3470.20$$ --- **Final answers:** - Annually: 3397.17 - Semiannually: 3433.18 - Quarterly: 3451.56 - Monthly: 3459.48 - Continuously: 3470.20