1. **Problem statement:** Find at least three different sequences starting with the terms 1, 2, 4, each generated by a simple formula. 2. **Approach:** We want formulas that generate sequences where the first three terms are 1, 2, and 4. 3. **Sequence 1: Geometric progression** - Formula: $a_n = a_1 \cdot r^{n-1}$ - Given $a_1=1$, $a_2=2$, so $2 = 1 \cdot r^{1} \Rightarrow r=2$ - Check $a_3 = 1 \cdot 2^{2} = 4$ correct. - So $a_n = 2^{n-1}$ 4. **Sequence 2: Quadratic sequence** - Assume $a_n = an^2 + bn + c$ - Use terms: - $a_1 = a + b + c = 1$ - $a_2 = 4a + 2b + c = 2$ - $a_3 = 9a + 3b + c = 4$ - Solve system: - From first: $c = 1 - a - b$ - Substitute into second: $4a + 2b + 1 - a - b = 2 \Rightarrow 3a + b = 1$ - Substitute into third: $9a + 3b + 1 - a - b = 4 \Rightarrow 8a + 2b = 3$ - From $3a + b = 1$, $b = 1 - 3a$ - Substitute into $8a + 2b = 3$: $8a + 2(1 - 3a) = 3 \Rightarrow 8a + 2 - 6a = 3 \Rightarrow 2a = 1 \Rightarrow a = \frac{1}{2}$ - Then $b = 1 - 3 \cdot \frac{1}{2} = 1 - \frac{3}{2} = -\frac{1}{2}$ - And $c = 1 - \frac{1}{2} - (-\frac{1}{2}) = 1$ - So $a_n = \frac{1}{2}n^2 - \frac{1}{2}n + 1$ 5. **Sequence 3: Linear recurrence** - Assume $a_n = p a_{n-1} + q a_{n-2}$ - Use $a_1=1$, $a_2=2$, $a_3=4$ - From $a_3 = p a_2 + q a_1$: $4 = 2p + q$ - Choose $p=1$, then $q=4 - 2(1) = 2$ - So recurrence: $a_n = a_{n-1} + 2 a_{n-2}$ with $a_1=1$, $a_2=2$ --- 6. **Now for the sequence starting with 3, 5, 7:** 7. **Sequence 1: Arithmetic progression** - Formula: $a_n = a_1 + (n-1)d$ - Given $a_1=3$, $a_2=5$, so $5 = 3 + d \Rightarrow d=2$ - Check $a_3 = 3 + 2 \cdot 2 = 7$ correct. - So $a_n = 3 + 2(n-1)$ 8. **Sequence 2: Quadratic sequence** - Assume $a_n = an^2 + bn + c$ - Use terms: - $a_1 = a + b + c = 3$ - $a_2 = 4a + 2b + c = 5$ - $a_3 = 9a + 3b + c = 7$ - Solve system: - $c = 3 - a - b$ - Substitute into second: $4a + 2b + 3 - a - b = 5 \Rightarrow 3a + b = 2$ - Substitute into third: $9a + 3b + 3 - a - b = 7 \Rightarrow 8a + 2b = 4$ - From $3a + b = 2$, $b = 2 - 3a$ - Substitute into $8a + 2b = 4$: $8a + 2(2 - 3a) = 4 \Rightarrow 8a + 4 - 6a = 4 \Rightarrow 2a = 0 \Rightarrow a = 0$ - Then $b = 2 - 0 = 2$ - And $c = 3 - 0 - 2 = 1$ - So $a_n = 2n + 1$ 9. **Sequence 3: Geometric progression with ratio 1** - Formula: $a_n = a_1 \cdot r^{n-1}$ - Given $a_1=3$, $a_2=5$, so $5 = 3r \Rightarrow r = \frac{5}{3}$ - Check $a_3 = 3 \cdot (\frac{5}{3})^2 = 3 \cdot \frac{25}{9} = \frac{75}{9} = \frac{25}{3} \neq 7$ - So geometric progression does not fit exactly. 10. **Alternative Sequence 3: Linear recurrence** - Assume $a_n = p a_{n-1} + q a_{n-2}$ - Use $a_1=3$, $a_2=5$, $a_3=7$ - From $a_3 = p a_2 + q a_1$: $7 = 5p + 3q$ - Choose $p=1$, then $q = \frac{7 - 5}{3} = \frac{2}{3}$ - So recurrence: $a_n = a_{n-1} + \frac{2}{3} a_{n-2}$ with $a_1=3$, $a_2=5$ **Final answers:** - For 1,2,4: 1) $a_n = 2^{n-1}$ 2) $a_n = \frac{1}{2}n^2 - \frac{1}{2}n + 1$ 3) $a_n = a_{n-1} + 2 a_{n-2}$ with $a_1=1$, $a_2=2$ - For 3,5,7: 1) $a_n = 3 + 2(n-1)$ 2) $a_n = 2n + 1$ 3) $a_n = a_{n-1} + \frac{2}{3} a_{n-2}$ with $a_1=3$, $a_2=5$