1. **Problem:** Use difference tables to find the next term in each sequence. **Sequence 1:** 2, 5, 8, 11, 14, ... - Step 1: Calculate first differences: 5-2=3, 8-5=3, 11-8=3, 14-11=3 - Step 2: Since first differences are constant (3), the sequence is arithmetic. - Step 3: Next term = last term + common difference = 14 + 3 = 17 **Sequence 2:** 5, 14, 27, 44, 65, ... - Step 1: First differences: 14-5=9, 27-14=13, 44-27=17, 65-44=21 - Step 2: Second differences: 13-9=4, 17-13=4, 21-17=4 (constant) - Step 3: Since second differences are constant, the sequence is quadratic. - Step 4: Next second difference = 4 - Step 5: Next first difference = last first difference + next second difference = 21 + 4 = 25 - Step 6: Next term = last term + next first difference = 65 + 25 = 90 **Sequence 3:** 2, 7, 24, 59, 118, 207, ... - Step 1: First differences: 7-2=5, 24-7=17, 59-24=35, 118-59=59, 207-118=89 - Step 2: Second differences: 17-5=12, 35-17=18, 59-35=24, 89-59=30 - Step 3: Third differences: 18-12=6, 24-18=6, 30-24=6 (constant) - Step 4: Since third differences are constant, the sequence is cubic. - Step 5: Next third difference = 6 - Step 6: Next second difference = last second difference + next third difference = 30 + 6 = 36 - Step 7: Next first difference = last first difference + next second difference = 89 + 36 = 125 - Step 8: Next term = last term + next first difference = 207 + 125 = 332 2. **Problem:** Find the nth-term formula for the number of square tiles in each figure sequence. **a) L-shape figures:** - Given terms: a_1=1, a_2=5, a_3=12, a_4=?, a_5=? - Step 1: Calculate differences: First differences: 5-1=4, 12-5=7 Second differences: 7-4=3 - Step 2: The pattern suggests quadratic formula: $a_n = An^2 + Bn + C$ - Step 3: Use known terms: For n=1: $A(1)^2 + B(1) + C = 1$ For n=2: $4A + 2B + C = 5$ For n=3: $9A + 3B + C = 12$ - Step 4: Solve system: From n=1: $A + B + C = 1$ From n=2: $4A + 2B + C = 5$ From n=3: $9A + 3B + C = 12$ - Step 5: Subtract equations: (4A+2B+C) - (A+B+C) = 5 - 1 => 3A + B = 4 (9A+3B+C) - (4A+2B+C) = 12 - 5 => 5A + B = 7 - Step 6: Subtract these: (5A + B) - (3A + B) = 7 - 4 => 2A = 3 => $A = \frac{3}{2}$ - Step 7: Substitute $A$ back: $3(\frac{3}{2}) + B = 4$ => $\frac{9}{2} + B = 4$ => $B = 4 - \frac{9}{2} = -\frac{1}{2}$ - Step 8: Find $C$: $A + B + C = 1$ => $\frac{3}{2} - \frac{1}{2} + C = 1$ => $1 + C = 1$ => $C=0$ - Step 9: Formula: $a_n = \frac{3}{2}n^2 - \frac{1}{2}n$ **b) Rectangular shapes:** - Given terms: 3, 6, 10, 15, 21 - Step 1: Recognize these as triangular numbers: $a_n = \frac{n(n+1)}{2} + 1$ or check differences - Step 2: Differences: 6-3=3, 10-6=4, 15-10=5, 21-15=6 - Step 3: Differences increase by 1, so formula is quadratic. - Step 4: Use $a_n = An^2 + Bn + C$ - Step 5: Use points: n=1: $A + B + C = 3$ n=2: $4A + 2B + C = 6$ n=3: $9A + 3B + C = 10$ - Step 6: Subtract: (4A+2B+C) - (A+B+C) = 6 - 3 => 3A + B = 3 (9A+3B+C) - (4A+2B+C) = 10 - 6 => 5A + B = 4 - Step 7: Subtract: (5A + B) - (3A + B) = 4 - 3 => 2A = 1 => $A=\frac{1}{2}$ - Step 8: Substitute $A$: $3(\frac{1}{2}) + B = 3$ => $\frac{3}{2} + B = 3$ => $B = \frac{3}{2}$ - Step 9: Find $C$: $\frac{1}{2} + \frac{3}{2} + C = 3$ => $2 + C = 3$ => $C=1$ - Step 10: Formula: $a_n = \frac{1}{2}n^2 + \frac{3}{2}n + 1$ **c) Stepped shapes:** - Given terms: 2, 5, 8, 12, 17 - Step 1: Differences: 5-2=3, 8-5=3, 12-8=4, 17-12=5 - Step 2: Differences are not constant, try quadratic formula. - Step 3: Use $a_n = An^2 + Bn + C$ - Step 4: Use points: n=1: $A + B + C = 2$ n=2: $4A + 2B + C = 5$ n=3: $9A + 3B + C = 8$ - Step 5: Subtract: (4A+2B+C) - (A+B+C) = 5 - 2 => 3A + B = 3 (9A+3B+C) - (4A+2B+C) = 8 - 5 => 5A + B = 3 - Step 6: Subtract: (5A + B) - (3A + B) = 3 - 3 => 2A = 0 => $A=0$ - Step 7: Substitute $A=0$: $3(0) + B = 3$ => $B=3$ - Step 8: Find $C$: $0 + 3 + C = 2$ => $C = -1$ - Step 9: Formula: $a_n = 3n - 1$ 3. **Problem:** You save 25 on day 1, and each day you save double the previous day. Find total savings after 4, 8, and 12 days. - Step 1: This is a geometric sequence with first term $a=25$ and common ratio $r=2$. - Step 2: Total savings after $n$ days is sum of geometric series: $$S_n = a \frac{r^n - 1}{r - 1}$$ - a) For 4 days: $$S_4 = 25 \frac{2^4 - 1}{2 - 1} = 25 \times (16 - 1) = 25 \times 15 = 375$$ - b) For 8 days: $$S_8 = 25 \frac{2^8 - 1}{1} = 25 \times (256 - 1) = 25 \times 255 = 6375$$ - c) For 12 days: $$S_{12} = 25 \times (2^{12} - 1) = 25 \times (4096 - 1) = 25 \times 4095 = 102375$$