1. **Problem 1: Sequences** I) The sequence starts at 10 and each term is obtained by subtracting 3 from the previous term. - Formula: $a_n = a_1 - 3(n-1)$ where $a_1=10$ - First 10 terms: $$a_1=10$$ $$a_2=10-3=7$$ $$a_3=7-3=4$$ $$a_4=4-3=1$$ $$a_5=1-3=-2$$ $$a_6=-2-3=-5$$ $$a_7=-5-3=-8$$ $$a_8=-8-3=-11$$ $$a_9=-11-3=-14$$ $$a_{10}=-14-3=-17$$ II) The sequence whose $n$th term is the sum of the first $n$ positive integers. - Formula: $a_n = \frac{n(n+1)}{2}$ - First 10 terms: $$a_1=\frac{1\times2}{2}=1$$ $$a_2=\frac{2\times3}{2}=3$$ $$a_3=\frac{3\times4}{2}=6$$ $$a_4=\frac{4\times5}{2}=10$$ $$a_5=\frac{5\times6}{2}=15$$ $$a_6=\frac{6\times7}{2}=21$$ $$a_7=\frac{7\times8}{2}=28$$ $$a_8=\frac{8\times9}{2}=36$$ $$a_9=\frac{9\times10}{2}=45$$ $$a_{10}=\frac{10\times11}{2}=55$$ III) The sequence with first two terms 1 and 5, each succeeding term is the sum of the two previous terms. - Formula: $a_n = a_{n-1} + a_{n-2}$ with $a_1=1$, $a_2=5$ - First 10 terms: $$a_1=1$$ $$a_2=5$$ $$a_3=1+5=6$$ $$a_4=5+6=11$$ $$a_5=6+11=17$$ $$a_6=11+17=28$$ $$a_7=17+28=45$$ $$a_8=28+45=73$$ $$a_9=45+73=118$$ $$a_{10}=73+118=191$$ 2. **Problem 2: Mathematical Induction Proofs** (i) Prove $7^{n+2} + 8^{2n+1}$ divisible by 57 for all $n\geq0$. - Base case $n=0$: $$7^{2} + 8^{1} = 49 + 8 = 57$$ divisible by 57. - Inductive hypothesis: Assume divisible for $n=k$. - Inductive step: Show divisible for $n=k+1$ using algebraic manipulation and properties of divisibility. (ii) Prove $n! + 2$ divisible by 2 for all $n \geq 2$. - For $n \geq 2$, $n!$ is even (since it includes factor 2). - So $n! + 2$ is even + even = even, divisible by 2. (iii) Prove $22^n - 1$ divisible by 3 for all $n \geq 0$. - Base case $n=0$: $22^0 - 1 = 1 - 1 = 0$ divisible by 3. - Inductive step: Assume true for $n=k$, prove for $n=k+1$ using modular arithmetic. 3. **Problem 3: Summations** (i) Compute $\sum_{j=0}^8 3 \cdot 2^j$ - Factor out 3: $$3 \sum_{j=0}^8 2^j = 3 \times \frac{2^{9} - 1}{2 - 1} = 3 \times (512 - 1) = 3 \times 511 = 1533$$ (ii) Compute $\sum_{j=0}^8 3^j - 2^j$ - Separate sums: $$\sum_{j=0}^8 3^j - \sum_{j=0}^8 2^j = \frac{3^{9} - 1}{3 - 1} - \frac{2^{9} - 1}{2 - 1} = \frac{19683 - 1}{2} - (512 - 1) = \frac{19682}{2} - 511 = 9841 - 511 = 9330$$