1. The binomial expansion of $(1 + px)^n$ starts with terms 1, $20x$, and $160x^2$. 2. The first term is always 1. 3. The second term is $npx = 20x$, so $np = 20$. 4. The third term is $\frac{n(n-1)}{2}p^2x^2 = 160x^2$, so $$\frac{n(n-1)}{2}p^2 = 160.$$ 5. We have two equations: $$np = 20$$ $$\frac{n(n-1)}{2}p^2 = 160.$$ 6. Express $p$ from the first: $p = \frac{20}{n}$. Substitute into second: $$\frac{n(n-1)}{2} \left(\frac{20}{n}\right)^2 = 160$$ $$\frac{n(n-1)}{2} \frac{400}{n^2} = 160$$ $$\frac{400(n-1)}{2n} = 160$$ $$\frac{400(n-1)}{2n} = 160$$ $$200 \frac{(n-1)}{n} = 160$$ $$\frac{n-1}{n} = \frac{160}{200} = 0.8$$ 7. Multiply both sides by $n$: $$n - 1 = 0.8n$$ $$n - 0.8n = 1$$ $$0.2n = 1$$ $$n = 5.$$ 8. From $np = 20$: $$5p = 20$$ $$p = 4.$$ 9. The values are $n = 5$ and $p = 4$. --- 10. Kwame's compound interest formula: $$CI = P(1 + \frac{r}{n})^{nt}$$ 11. Given principal $P = 5000$, rate $r = 0.05$ (5%), number of times compounded per year $n = 1$ (assuming annually), time $t = 3$ years. 12. Calculate: $$CI = 5000\times (1 + \frac{0.05}{1})^{1\times 3} = 5000 \times (1.05)^3.$$ 13. Use binomial theorem to expand $(1.05)^3$: $$(1 + 0.05)^3 = 1 + 3(0.05) + 3(0.05)^2 + (0.05)^3 = 1 + 0.15 + 0.0075 + 0.000125 = 1.157625.$$ 14. Multiply to find total amount: $$5000 \times 1.157625 = 5788.125.$$ 15. Thus Kwame pays $5788.125$ at the end of 3 years. --- 16. To find the $x^5$ term in the expansion of $(2x - y)^7$, the general term is $$T_{r+1} = \binom{7}{r}(2x)^{7-r}(-y)^r.$$ 17. The term with $x^5$ means the power on $x$ is 5, so $$7 - r = 5 \implies r = 2.$$ 18. Substitute $r=2$: $$T_3 = \binom{7}{2} (2x)^5 (-y)^2 = \binom{7}{2} 2^5 x^5 y^2.$$ 19. Calculate: $$\binom{7}{2} = \frac{7\times 6}{2} = 21,$$ $$2^5 = 32.$$ 20. Thus, $$T_3 = 21 \times 32 x^5 y^2 = 672 x^5 y^2.$$