1. **Stating the problem:** We are given the sequence 2, 8, 18, 32, 50 and need to find the formula for the nth term. 2. **Look for a pattern:** Let's examine the differences between terms: - $8 - 2 = 6$ - $18 - 8 = 10$ - $32 - 18 = 14$ - $50 - 32 = 18$ 3. **Second differences:** Now find the differences of these differences: - $10 - 6 = 4$ - $14 - 10 = 4$ - $18 - 14 = 4$ Since the second differences are constant (4), the sequence is quadratic and the nth term has the form: $$a_n = An^2 + Bn + C$$ 4. **Set up equations using the first three terms:** - For $n=1$: $A(1)^2 + B(1) + C = 2$ - For $n=2$: $A(2)^2 + B(2) + C = 8$ - For $n=3$: $A(3)^2 + B(3) + C = 18$ This gives: - $A + B + C = 2$ - $4A + 2B + C = 8$ - $9A + 3B + C = 18$ 5. **Solve the system:** Subtract the first equation from the second: $$ (4A + 2B + C) - (A + B + C) = 8 - 2 \Rightarrow 3A + B = 6 $$ Subtract the second from the third: $$ (9A + 3B + C) - (4A + 2B + C) = 18 - 8 \Rightarrow 5A + B = 10 $$ Subtract the first new equation from the second: $$ (5A + B) - (3A + B) = 10 - 6 \Rightarrow 2A = 4 \Rightarrow A = 2 $$ Substitute $A=2$ into $3A + B = 6$: $$ 3(2) + B = 6 \Rightarrow 6 + B = 6 \Rightarrow B = 0 $$ Substitute $A=2$, $B=0$ into $A + B + C = 2$: $$ 2 + 0 + C = 2 \Rightarrow C = 0 $$ 6. **Final formula:** $$a_n = 2n^2$$ 7. **Verification:** - $a_1 = 2(1)^2 = 2$ - $a_2 = 2(2)^2 = 8$ - $a_3 = 2(3)^2 = 18$ - $a_4 = 2(4)^2 = 32$ - $a_5 = 2(5)^2 = 50$ All terms match the given sequence. **Answer:** The nth term is $$a_n = 2n^2$$.