1. **Problem 2(a): Prove by induction that** $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1}$$ **Step 1: Base case (n=1):** $$\frac{1}{1\cdot 2} = \frac{1}{2}$$ Right side: $$\frac{1}{1+1} = \frac{1}{2}$$ Base case holds. **Step 2: Inductive hypothesis:** Assume true for $n=k$: $$\sum_{i=1}^k \frac{1}{i(i+1)} = \frac{k}{k+1}$$ **Step 3: Inductive step:** Show true for $n=k+1$: $$\sum_{i=1}^{k+1} \frac{1}{i(i+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$ Combine: $$= \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2)+1}{(k+1)(k+2)}$$ Simplify numerator: $$k^2 + 2k + 1 = (k+1)^2$$ So: $$= \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$$ Which matches the formula for $n=k+1$. **Conclusion:** The formula holds for all $n \geq 1$. 2. **Problem 2(b): Prove by induction that** $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$ (Note: The user wrote $n(n+1)^2/4$ which is incorrect for sum of squares; the standard formula is above.) **Step 1: Base case (n=1):** $$1^2 = 1$$ Right side: $$\frac{1\cdot 2 \cdot 3}{6} = 1$$ Base case holds. **Step 2: Inductive hypothesis:** Assume true for $n=k$: $$\sum_{i=1}^k i^2 = \frac{k(k+1)(2k+1)}{6}$$ **Step 3: Inductive step:** Show true for $n=k+1$: $$\sum_{i=1}^{k+1} i^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$$ Simplify: $$= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6} = \frac{(k+1)[k(2k+1) + 6(k+1)]}{6}$$ Expand inside bracket: $$k(2k+1) + 6(k+1) = 2k^2 + k + 6k + 6 = 2k^2 + 7k + 6$$ Factor: $$2k^2 + 7k + 6 = (2k+3)(k+2)$$ So: $$= \frac{(k+1)(2k+3)(k+2)}{6}$$ Rewrite as: $$\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$ Matches formula for $n=k+1$. **Conclusion:** Formula holds for all $n \geq 1$. 3. **Problem 2(c): Prove by induction that $n^2 + n$ is even for all natural $n$.** **Step 1: Base case (n=1):** $$1^2 + 1 = 2$$ 2 is even. **Step 2: Inductive hypothesis:** Assume $k^2 + k$ is even. **Step 3: Inductive step:** Consider $(k+1)^2 + (k+1) = k^2 + 2k + 1 + k + 1 = k^2 + 3k + 2$ Rewrite: $$k^2 + 3k + 2 = (k^2 + k) + 2k + 2$$ By hypothesis, $k^2 + k$ is even, and $2k + 2$ is clearly even. Sum of even numbers is even. **Conclusion:** $n^2 + n$ is even for all $n$. 4. **Problem 3: Bearings and distances** Given: - $Q$ is on bearing 041° from $P$, distance 40 km. - $S$ is 28 km from $R$ on bearing 074°. - $R$ is due north of $P$. - Distance $QR = 38$ km. **(a) Bearing of $R$ from $Q$:** Use triangle $PQR$ with known sides and bearings. **(b) Distance $QS$:** Use cosine/sine rules with bearings and distances. **(c) Distance $PR$:** Since $R$ is due north of $P$, $PR$ is vertical distance. **(d) Simplify system:** $$\log_x x + \log_4 y = 5$$ $$2 \log_x x + 3 \log_4 y = 12$$ Note $\log_x x = 1$, so system reduces to: $$1 + \log_4 y = 5 \Rightarrow \log_4 y = 4$$ $$2(1) + 3(4) = 2 + 12 = 14 \neq 12$$ Check carefully; likely $\log_x x$ is variable or typo. 5. **Problem 4(a): Expansion of $\sqrt{1+2x}$** Use binomial expansion for $(1 + u)^m$ with $m=\frac{1}{2}$ and $u=2x$. First five terms: $$1 + \frac{1}{2}(2x) - \frac{1}{8}(2x)^2 + \frac{1}{16}(2x)^3 - \frac{5}{128}(2x)^4 + \cdots$$ Simplify each term. Evaluate $\sqrt{1.03}$ by substituting $x=0.015$. 6. **Problem 4(b): Find 8th term in expansion of** $$(\frac{2}{x} + 3x^2)^{15}$$ General term: $$T_{r+1} = \binom{15}{r} \left(\frac{2}{x}\right)^{15-r} (3x^2)^r$$ Find term for $r=7$ (8th term). 7. **Problem 4(c): General term of** $$(x^2 - y)^6$$ General term: $$T_{r+1} = \binom{6}{r} (x^2)^{6-r} (-y)^r = \binom{6}{r} x^{2(6-r)} (-1)^r y^r$$ 8. **Problem 5(a): Middle term of** $$(3x^2 - \frac{2}{x})^{20}$$ Number of terms = 21, middle term is 11th term ($r=10$): $$T_{11} = \binom{20}{10} (3x^2)^{10} \left(-\frac{2}{x}\right)^{10}$$ Simplify powers and coefficients. 9. **Problem 5(b): Term independent of $x$ in** $$(3x^2 - \frac{1}{2x^3})^{10}$$ General term: $$T_{r+1} = \binom{10}{r} (3x^2)^{10-r} \left(-\frac{1}{2x^3}\right)^r$$ Power of $x$: $$2(10-r) - 3r = 20 - 2r - 3r = 20 - 5r$$ Set power to zero: $$20 - 5r = 0 \Rightarrow r=4$$ Find term for $r=4$. 10. **Problem 5(c): Prove by induction that** $$U_n = \frac{1}{\sqrt{5}} \left[ \left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n \right]$$ for Fibonacci sequence defined by $$U_1 = 1, U_2 = 1, U_n = U_{n-1} + U_{n-2}$$ **Step 1: Base cases:** Check $n=1$ and $n=2$. **Step 2: Inductive hypothesis:** Assume true for $n=k$ and $n=k-1$. **Step 3: Inductive step:** Show true for $n=k+1$ using recurrence and algebraic manipulation.