1. **Problem 21a:** Find the binomial expansion of $\left(1 - \frac{x}{4}\right)^5$ in ascending powers of $x$. Using the binomial theorem: $$\left(1 - \frac{x}{4}\right)^5 = \sum_{k=0}^5 \binom{5}{k} 1^{5-k} \left(-\frac{x}{4}\right)^k = \sum_{k=0}^5 \binom{5}{k} (-1)^k \frac{x^k}{4^k}$$ Calculate terms: - $k=0$: $\binom{5}{0} (-1)^0 \frac{x^0}{4^0} = 1$ - $k=1$: $\binom{5}{1} (-1)^1 \frac{x}{4} = 5 \times (-1) \times \frac{x}{4} = -\frac{5x}{4}$ - $k=2$: $\binom{5}{2} (-1)^2 \frac{x^2}{4^2} = 10 \times 1 \times \frac{x^2}{16} = \frac{10x^2}{16} = \frac{5x^2}{8}$ - $k=3$: $\binom{5}{3} (-1)^3 \frac{x^3}{4^3} = 10 \times (-1) \times \frac{x^3}{64} = -\frac{10x^3}{64} = -\frac{5x^3}{32}$ - $k=4$: $\binom{5}{4} (-1)^4 \frac{x^4}{4^4} = 5 \times 1 \times \frac{x^4}{256} = \frac{5x^4}{256}$ - $k=5$: $\binom{5}{5} (-1)^5 \frac{x^5}{4^5} = 1 \times (-1) \times \frac{x^5}{1024} = -\frac{x^5}{1024}$ So the expansion is: $$1 - \frac{5x}{4} + \frac{5x^2}{8} - \frac{5x^3}{32} + \frac{5x^4}{256} - \frac{x^5}{1024}$$ 2. **Problem 21b:** Using the first three terms, approximate $0.975^5$. Note that $0.975 = 1 - 0.025$, so $x = 0.1$ because $\frac{x}{4} = 0.025 \Rightarrow x = 0.1$. Using first three terms: $$1 - \frac{5(0.1)}{4} + \frac{5(0.1)^2}{8} = 1 - \frac{0.5}{4} + \frac{5 \times 0.01}{8} = 1 - 0.125 + 0.00625 = 0.88125$$ 3. **Problem 22a:** Arithmetic series with 15th term $a_{15} = 143$ and 31st term $a_{31} = 183$. General term: $a_n = a_1 + (n-1)d$ Set up equations: $$a_1 + 14d = 143$$ $$a_1 + 30d = 183$$ Subtract first from second: $$16d = 40 \Rightarrow d = \frac{40}{16} = 2.5$$ Find $a_1$: $$a_1 = 143 - 14 \times 2.5 = 143 - 35 = 108$$ 4. **Problem 22b:** Find 100th term: $$a_{100} = a_1 + 99d = 108 + 99 \times 2.5 = 108 + 247.5 = 355.5$$ 5. **Problem 23a:** Compound interest on $3000$ at 1.5% per year for 10 years. Formula: $$A = P(1 + r)^t = 3000 (1 + 0.015)^{10}$$ Calculate: $$A = 3000 \times (1.015)^{10} \approx 3000 \times 1.160968 = 3482.9$$ Interest earned: $$3482.9 - 3000 = 482.9 \approx 483$$ 6. **Problem 23b:** Additional $1200$ deposited annually starting 1 Jan 2020 for 10 years total (deposits at start of each year except 2019). Calculate total amount at start of Jan 2030 before deposit that year. - Initial $3000$ grows 11 years (2019 to 2030): $$3000 (1.015)^{11} \approx 3000 \times 1.178 = 3534$$ - Annual deposits form a geometric series with first term $1200 (1.015)^{10}$ (deposited at start of 2020, grows 10 years), ratio $1.015$, number of terms $10$. Sum of deposits: $$S = 1200 \times \frac{(1.015^{10} - 1)}{1.015 - 1} = 1200 \times \frac{1.160968 - 1}{0.015} = 1200 \times \frac{0.160968}{0.015} = 1200 \times 10.7312 = 12877.44$$ Total amount: $$3534 + 12877.44 = 16411.44$$ 7. **Problem 24a:** Brad deposits $5500$ at 2.75% compound interest for 4 years. $$A = 5500 (1 + 0.0275)^4 = 5500 (1.0275)^4$$ Calculate: $$(1.0275)^4 \approx 1.1145$$ $$A \approx 5500 \times 1.1145 = 6130$$ 8. **Problem 24b:** Find years $t$ for investment to reach $12000$. $$12000 = 5500 (1.0275)^t$$ $$\Rightarrow (1.0275)^t = \frac{12000}{5500} = 2.1818$$ Take natural logs: $$t \ln(1.0275) = \ln(2.1818)$$ $$t = \frac{\ln(2.1818)}{\ln(1.0275)} \approx \frac{0.780}{0.0271} = 28.8$$ Rounded to nearest year: $29$ years. 9. **Problem 25:** Coefficient of $x^5$ in $(3 + x)(4 + 2x)^8$. Expand $(4 + 2x)^8$ using binomial theorem: $$\sum_{k=0}^8 \binom{8}{k} 4^{8-k} (2x)^k = \sum_{k=0}^8 \binom{8}{k} 4^{8-k} 2^k x^k$$ Multiply by $(3 + x)$: $$3 \sum_{k=0}^8 c_k x^k + x \sum_{k=0}^8 c_k x^k = \sum_{k=0}^8 3 c_k x^k + \sum_{k=0}^8 c_k x^{k+1}$$ Coefficient of $x^5$ comes from: - $3 c_5$ (from $3 \times x^5$ term) - $c_4$ (from $x \times x^4$ term) Calculate $c_k = \binom{8}{k} 4^{8-k} 2^k$: - $c_4 = \binom{8}{4} 4^{4} 2^{4} = 70 \times 256 \times 16 = 286720$ - $c_5 = \binom{8}{5} 4^{3} 2^{5} = 56 \times 64 \times 32 = 114688$ Coefficient of $x^5$: $$3 \times 114688 + 286720 = 344064 + 286720 = 630784$$ 10. **Problem 26:** Coefficient of $x^2$ in $(1 + 3x)^n$ is 495. Coefficient of $x^2$ is: $$\binom{n}{2} 3^2 = 9 \frac{n(n-1)}{2} = 495$$ Solve: $$\frac{9 n (n-1)}{2} = 495 \Rightarrow 9 n (n-1) = 990 \Rightarrow n (n-1) = 110$$ Solve quadratic: $$n^2 - n - 110 = 0$$ $$n = \frac{1 \pm \sqrt{1 + 440}}{2} = \frac{1 \pm 21}{2}$$ Positive root: $$n = \frac{1 + 21}{2} = 11$$ 11. **Problem 27:** Constant term in $(x^3 - \frac{2}{x})^8$. General term: $$\binom{8}{k} (x^3)^k \left(-\frac{2}{x}\right)^{8-k} = \binom{8}{k} (-1)^{8-k} 2^{8-k} x^{3k} x^{-(8-k)} = \binom{8}{k} (-1)^{8-k} 2^{8-k} x^{3k - 8 + k} = \binom{8}{k} (-1)^{8-k} 2^{8-k} x^{4k - 8}$$ Constant term when exponent of $x$ is zero: $$4k - 8 = 0 \Rightarrow k = 2$$ Calculate term for $k=2$: $$\binom{8}{2} (-1)^6 2^6 = 28 \times 1 \times 64 = 1792$$ 12. **Problem 28a:** Binomial expansion of $\left(\frac{1}{2}x - x\right)^4$ in ascending powers of $x$. Rewrite: $$\left(\frac{1}{2}x - x\right)^4 = \left(-\frac{1}{2}x\right)^4 = \left(-\frac{1}{2} + 1\right)^4 x^4$$ But better to expand directly: $$(\frac{1}{2}x - x)^4 = x^4 \left(\frac{1}{2} - 1\right)^4 = x^4 \left(-\frac{1}{2}\right)^4 = x^4 \times \frac{1}{16} = \frac{x^4}{16}$$ Alternatively, expand using binomial theorem: $$\sum_{k=0}^4 \binom{4}{k} \left(\frac{1}{2}x\right)^k (-x)^{4-k} = \sum_{k=0}^4 \binom{4}{k} \left(\frac{1}{2}\right)^k (-1)^{4-k} x^{k + 4 - k} = x^4 \sum_{k=0}^4 \binom{4}{k} \left(\frac{1}{2}\right)^k (-1)^{4-k}$$ Calculate sum: $$\sum_{k=0}^4 \binom{4}{k} \left(\frac{1}{2}\right)^k (-1)^{4-k} = (\text{expand and sum}) = \frac{1}{16}$$ So expansion is: $$\frac{x^4}{16}$$ 13. **Problem 28b:** Find term independent of $x$ in $(3 - x)^3 \left(\frac{1}{2}x - x\right)^4$. From above, $(\frac{1}{2}x - x)^4 = \frac{x^4}{16}$. Expand $(3 - x)^3$: $$\sum_{m=0}^3 \binom{3}{m} 3^{3-m} (-x)^m = \sum_{m=0}^3 \binom{3}{m} 3^{3-m} (-1)^m x^m$$ Multiply by $\frac{x^4}{16}$: $$\sum_{m=0}^3 \binom{3}{m} 3^{3-m} (-1)^m \frac{x^{m+4}}{16}$$ Term independent of $x$ means $m+4=0$ which is impossible since $m \geq 0$. Hence, no term independent of $x$ from this product. But question likely means find term independent of $x$ in the product, so consider the full expansion carefully. Alternatively, rewrite $(\frac{1}{2}x - x)^4 = x^4 (\frac{1}{2} - 1)^4 = x^4 (-\frac{1}{2})^4 = \frac{x^4}{16}$. So the product is: $$(3 - x)^3 \times \frac{x^4}{16}$$ The lowest power of $x$ is 4, so no constant term. If question means find term independent of $x$ in the expansion of $(3 - x)^3 (\frac{1}{2}x - x)^4$ considering the binomial expansion of $(\frac{1}{2}x - x)^4$ as separate terms, then the term independent of $x$ comes from the product of the term in $(3 - x)^3$ with $x^0$ and the term in $(\frac{1}{2}x - x)^4$ with $x^0$. Since $(\frac{1}{2}x - x)^4$ has no constant term, the constant term in the product is zero. 14. **Problem 29:** Geometric series with sum to infinity $S_\infty = 120$, common ratio $r=0.2$. Sum to infinity formula: $$S_\infty = \frac{a}{1-r} = 120 \Rightarrow a = 120 (1 - 0.2) = 120 \times 0.8 = 96$$ Find 6th term: $$t_6 = a r^{5} = 96 \times (0.2)^5 = 96 \times 0.00032 = 0.03072$$ **Summary:** - 21a: $1 - \frac{5x}{4} + \frac{5x^2}{8} - \frac{5x^3}{32} + \frac{5x^4}{256} - \frac{x^5}{1024}$ - 21b: Approx $0.88125$ - 22a: $a_1=108$, $d=2.5$ - 22b: $a_{100}=355.5$ - 23a: Interest $483$ - 23b: Total $16411$ - 24a: Value $6130$ - 24b: Years $29$ - 25: Coefficient $630784$ - 26: $n=11$ - 27: Constant term $1792$ - 28a: $\frac{x^4}{16}$ - 28b: No constant term - 29: 6th term $0.03072$