1. **Stating the problem:** We are given the sequence of numbers 4, 6, 10 and asked to find the pattern. 2. **Observing the sequence:** The numbers are 4, 6, 10. 3. **Check the differences between consecutive terms:** $$6 - 4 = 2$$ $$10 - 6 = 4$$ 4. **Check the second difference:** $$4 - 2 = 2$$ 5. Since the second difference is constant (2), this suggests the sequence follows a quadratic pattern. 6. **General form of a quadratic sequence:** $$a_n = An^2 + Bn + C$$ 7. **Use the first three terms to find A, B, and C:** For $n=1$, $a_1 = A(1)^2 + B(1) + C = A + B + C = 4$ For $n=2$, $a_2 = 4A + 2B + C = 6$ For $n=3$, $a_3 = 9A + 3B + C = 10$ 8. **Set up the system of equations:** $$\begin{cases} A + B + C = 4 \\ 4A + 2B + C = 6 \\ 9A + 3B + C = 10 \end{cases}$$ 9. **Subtract the first equation from the second and third:** $$ (4A + 2B + C) - (A + B + C) = 6 - 4 \Rightarrow 3A + B = 2 $$ $$ (9A + 3B + C) - (A + B + C) = 10 - 4 \Rightarrow 8A + 2B = 6 $$ 10. **Simplify the second equation:** $$8A + 2B = 6 \Rightarrow 4A + B = 3$$ 11. **Now solve the system:** $$\begin{cases} 3A + B = 2 \\ 4A + B = 3 \end{cases}$$ 12. **Subtract the first from the second:** $$ (4A + B) - (3A + B) = 3 - 2 \Rightarrow A = 1 $$ 13. **Substitute $A=1$ into $3A + B = 2$:** $$3(1) + B = 2 \Rightarrow B = -1$$ 14. **Substitute $A=1$, $B=-1$ into $A + B + C = 4$:** $$1 - 1 + C = 4 \Rightarrow C = 4$$ 15. **Therefore, the formula for the $n$th term is:** $$a_n = n^2 - n + 4$$ 16. **Check the formula:** For $n=1$: $1 - 1 + 4 = 4$ For $n=2$: $4 - 2 + 4 = 6$ For $n=3$: $9 - 3 + 4 = 10$ **Final answer:** The pattern follows the quadratic formula $$a_n = n^2 - n + 4$$.