1. The problem asks to determine whether a figure with 211 dots exists in the sequence and explain why. 2. From the previous parts (implied), the nth term formula for the number of dots is needed. Observing the pattern of dots: Figure 1 has 4 dots, Figure 2 has 6 dots, Figure 3 has 9 dots, Figure 4 has 12 dots. 3. The differences between terms are: 6 - 4 = 2, 9 - 6 = 3, 12 - 9 = 3. The pattern suggests the nth term formula is linear or quadratic. Checking the pattern, the nth term is given by: $$a_n = 3n + 1$$ (assuming from the pattern: 1st term 4 = 3(1)+1, 2nd term 6 = 3(2)+0? Actually 3(2)+1=7, so this is not correct. Let's try another approach.) 4. Alternatively, the sequence is 4, 6, 9, 12. Differences: 2, 3, 3. The second difference is 1 (3-2=1, 3-3=0), so possibly quadratic. Try quadratic formula: $$a_n = An^2 + Bn + C$$ Using points: For n=1, a_1=4: $$A(1)^2 + B(1) + C = 4$$ For n=2, a_2=6: $$4A + 2B + C = 6$$ For n=3, a_3=9: $$9A + 3B + C = 9$$ Subtract first from second: $$4A + 2B + C - (A + B + C) = 6 - 4$$ $$3A + B = 2$$ Subtract first from third: $$9A + 3B + C - (A + B + C) = 9 - 4$$ $$8A + 2B = 5$$ Multiply first by 2: $$6A + 2B = 4$$ Subtract from second: $$(8A + 2B) - (6A + 2B) = 5 - 4$$ $$2A = 1 ightarrow A = \frac{1}{2}$$ Substitute back: $$3(\frac{1}{2}) + B = 2 ightarrow \frac{3}{2} + B = 2 ightarrow B = \frac{1}{2}$$ Use first equation: $$\frac{1}{2} + \frac{1}{2} + C = 4 ightarrow 1 + C = 4 ightarrow C = 3$$ So the formula is: $$a_n = \frac{1}{2}n^2 + \frac{1}{2}n + 3$$ 5. To check if a figure with 211 dots exists, solve for n: $$\frac{1}{2}n^2 + \frac{1}{2}n + 3 = 211$$ Multiply both sides by 2: $$n^2 + n + 6 = 422$$ Simplify: $$n^2 + n - 416 = 0$$ 6. Solve quadratic equation: $$n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-416)}}{2} = \frac{-1 \pm \sqrt{1 + 1664}}{2} = \frac{-1 \pm \sqrt{1665}}{2}$$ 7. Since $\sqrt{1665}$ is not a perfect square, $n$ is not an integer. 8. Therefore, no figure with exactly 211 dots exists in the sequence because $n$ must be a positive integer and the equation has no integer solution. Final answer: No, a figure with 211 dots does not exist in the sequence because the quadratic equation for $n$ does not yield an integer solution.