1. **Problem statement:** The first five terms of a sequence are 4, 6, 16, 40, and 81. (i) Determine if the sequence is quadratic or cubic. (ii) Find a formula for the general term in terms of $n$. (iii) Find the 21st term. 2. **Step 1: Identify the type of sequence by differences.** Calculate the first differences: $$6-4=2, \quad 16-6=10, \quad 40-16=24, \quad 81-40=41$$ Calculate the second differences: $$10-2=8, \quad 24-10=14, \quad 41-24=17$$ Calculate the third differences: $$14-8=6, \quad 17-14=3$$ Since the third differences are not constant, check if the sequence is quadratic or cubic. 3. **Step 2: Check if quadratic or cubic.** For quadratic sequences, the second differences are constant. Here, second differences are 8, 14, 17 (not constant). For cubic sequences, the third differences should be constant. Here, third differences are 6 and 3 (not constant). Neither second nor third differences are constant, but the third differences are closer to constant. 4. **Step 3: Try to fit a cubic polynomial:** Assume general term: $$T_n = an^3 + bn^2 + cn + d$$ Use the first four terms to form equations: For $n=1$, $T_1=4$: $$a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d = 4$$ For $n=2$, $T_2=6$: $$8a + 4b + 2c + d = 6$$ For $n=3$, $T_3=16$: $$27a + 9b + 3c + d = 16$$ For $n=4$, $T_4=40$: $$64a + 16b + 4c + d = 40$$ 5. **Step 4: Solve the system of equations:** Subtract equation 1 from equations 2, 3, and 4: Equation 2 - Equation 1: $$7a + 3b + c = 2$$ Equation 3 - Equation 1: $$26a + 8b + 2c = 12$$ Equation 4 - Equation 1: $$63a + 15b + 3c = 36$$ 6. **Step 5: Simplify and solve for $a$, $b$, and $c$.** From equation 1 of this system: $$c = 2 - 7a - 3b$$ Substitute $c$ into the other two: Equation 2: $$26a + 8b + 2(2 - 7a - 3b) = 12$$ $$26a + 8b + 4 - 14a - 6b = 12$$ $$12a + 2b + 4 = 12$$ $$12a + 2b = 8$$ Equation 3: $$63a + 15b + 3(2 - 7a - 3b) = 36$$ $$63a + 15b + 6 - 21a - 9b = 36$$ $$42a + 6b + 6 = 36$$ $$42a + 6b = 30$$ 7. **Step 6: Solve the two equations:** $$12a + 2b = 8$$ $$42a + 6b = 30$$ Divide second equation by 3: $$14a + 2b = 10$$ Subtract first equation from this: $$(14a + 2b) - (12a + 2b) = 10 - 8$$ $$2a = 2$$ $$a = 1$$ 8. **Step 7: Find $b$ and $c$.** Substitute $a=1$ into $12a + 2b = 8$: $$12(1) + 2b = 8$$ $$12 + 2b = 8$$ $$2b = 8 - 12 = -4$$ $$b = -2$$ Substitute $a=1$, $b=-2$ into $c = 2 - 7a - 3b$: $$c = 2 - 7(1) - 3(-2) = 2 - 7 + 6 = 1$$ 9. **Step 8: Find $d$ using equation 1:** $$a + b + c + d = 4$$ $$1 - 2 + 1 + d = 4$$ $$0 + d = 4$$ $$d = 4$$ 10. **Step 9: Write the general term:** $$T_n = n^3 - 2n^2 + n + 4$$ 11. **Step 10: Find the 21st term:** $$T_{21} = 21^3 - 2(21)^2 + 21 + 4$$ Calculate: $$21^3 = 9261$$ $$21^2 = 441$$ $$T_{21} = 9261 - 2(441) + 21 + 4 = 9261 - 882 + 21 + 4 = 9404$$ **Final answers:** (i) The sequence is cubic. (ii) The general term is $$T_n = n^3 - 2n^2 + n + 4$$ (iii) The 21st term is $$9404$$.