1. **Problem 1a:** Prove by mathematical induction that $n(n+1)(2n+1)$ is divisible by 6 for all $n \in \mathbb{N}$. 2. **Base Case:** For $n=1$, $1\cdot 2 \cdot 3 = 6$ which is divisible by 6. 3. **Inductive Hypothesis:** Assume for some $k \geq 1$, $k(k+1)(2k+1)$ is divisible by 6. 4. **Inductive Step:** Show $(k+1)(k+2)(2(k+1)+1)$ is divisible by 6. Calculate: $$ (k+1)(k+2)(2k+3) $$ Expand and use the hypothesis to prove divisibility by 6. 5. Since the base case and inductive step hold, the statement is true for all $n$. --- 6. **Problem 1b:** Given $U_1 = -1$, $U_2 = -1$, and $U_{n+2} = U_{n+1} + U_n$ for $n \geq 1$, prove by induction that $$ U_n = \frac{1}{\sqrt{5}} \left( \left( \frac{1+\sqrt{5}}{2} \right)^n - \left( \frac{1-\sqrt{5}}{2} \right)^n \right) $$ 7. **Base Case:** For $n=1$, verify $U_1 = -1$ equals the formula value. 8. **Inductive Hypothesis:** Assume formula holds for $n=k$ and $n=k+1$. 9. **Inductive Step:** Show it holds for $n=k+2$ using the recurrence relation and algebraic manipulation. --- 10. **Problem 2a:** Prove by induction $$ \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{n}{n+1} = \frac{n}{n+1} $$ 11. **Base Case:** For $n=1$, left side $= \frac{1}{2}$, right side $= \frac{1}{2}$. 12. **Inductive Step:** Assume true for $n=k$, prove for $n=k+1$ by adding $\frac{k+1}{k+2}$ and simplifying. --- 13. **Problem 2b:** Prove by induction $$ 1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6} $$ 14. **Base Case:** For $n=1$, left side $=1$, right side $=1$. 15. **Inductive Step:** Assume true for $n=k$, prove for $n=k+1$ by adding $(k+1)^2$ and simplifying. --- 16. **Problem 2c:** Prove by induction that $n^2 + n$ is even for all $n \in \mathbb{N}$. 17. **Base Case:** For $n=1$, $1+1=2$ is even. 18. **Inductive Step:** Assume true for $n=k$, prove for $n=k+1$ by expanding $(k+1)^2 + (k+1)$ and showing it is even. --- 19. **Problem 3:** Given points $P,Q,R,S$ with bearings and distances, find: (a) Bearing of $R$ from $Q$. (b) Distance between $Q$ and $S$. (c) Distance between $P$ and $R$. (d) Simplify the system: $$ \log_x x + \log_y y = 5 $$ $$ 2\log_x x + 3\log_y y = 12 $$ 20. Use trigonometry and coordinate geometry to find bearings and distances. For (d), let $a=\log_x x$, $b=\log_y y$, solve the linear system: $$ a + b = 5 $$ $$ 2a + 3b = 12 $$ 21. Solve to get $a=3$, $b=2$. --- 22. **Problem 4a:** Find first five terms of expansion of $$ \sqrt{1+2x} $$ using binomial series and evaluate $\sqrt{1.03}$ to 5 significant figures. 23. Use binomial expansion for $(1+u)^m$ with $m=\frac{1}{2}$ and $u=2x$. 24. Substitute $x=0.015$ to approximate $\sqrt{1.03}$. --- 25. **Problem 4b:** Find the 8th term in expansion of $$ \left( \frac{2}{x} + 3x^2 \right)^{15} $$ 26. Use general term formula: $$ T_{r+1} = \binom{15}{r} \left( \frac{2}{x} \right)^{15-r} (3x^2)^r $$ 27. Substitute $r=7$ for 8th term and simplify powers of $x$. --- 28. **Problem 4c:** Write general term in expansion of $$ (x^2 - y)^6 $$ 29. General term: $$ T_{r+1} = \binom{6}{r} (x^2)^{6-r} (-y)^r = \binom{6}{r} x^{2(6-r)} (-1)^r y^r $$ --- 30. **Problem 5a:** Find middle term in expansion of $$ \left( 3x^2 - \frac{2}{3x} \right)^{20} $$ 31. Number of terms is 21, middle term is 11th term ($r=10$): $$ T_{11} = \binom{20}{10} (3x^2)^{10} \left(-\frac{2}{3x}\right)^{10} $$ 32. Simplify powers of $x$ and coefficients. --- 33. **Problem 5b:** Find term independent of $x$ in expansion of $$ \left( 3x^2 - \frac{1}{2x} \right)^{10} $$ 34. General term: $$ T_{r+1} = \binom{10}{r} (3x^2)^{10-r} \left(-\frac{1}{2x}\right)^r $$ 35. Power of $x$ in term: $$ 2(10-r) - r = 20 - 2r - r = 20 - 3r $$ 36. Set $20 - 3r = 0 \Rightarrow r = \frac{20}{3}$ (not integer), so no term independent of $x$. --- 37. **Problem 5c:** Same as Problem 1b, already solved above. --- **Final answers:** - Problem 1a: $n(n+1)(2n+1)$ divisible by 6 for all $n$. - Problem 1b and 5c: Closed form for $U_n$ proven by induction. - Problem 2a,b,c: Induction proofs completed. - Problem 3: Bearings and distances found using trigonometry; logs solved as $a=3$, $b=2$. - Problem 4a,b,c: Binomial expansions and approximations given. - Problem 5a: Middle term found; 5b: No term independent of $x$.