Nth Term Sequences 189478

1. **State the problem:** We have a base sequence with nth term $n^2$: 1, 4, 9, 16, ... 2. **Given sequences:** a) 2, 5, 10, 17, ... b) 2, 8, 18, 32, ... 3. **Goal:** Find the nth term formula for each sequence using the fact that the base sequence's nth term is $n^2$. --- ### Part a) 4. Compare each term of sequence a) with $n^2$: - For $n=1$: $n^2=1$, sequence a) term = 2 - For $n=2$: $n^2=4$, sequence a) term = 5 - For $n=3$: $n^2=9$, sequence a) term = 10 - For $n=4$: $n^2=16$, sequence a) term = 17 5. Notice each term in sequence a) is exactly 1 more than $n^2$. 6. Therefore, the nth term for sequence a) is: $$a_n = n^2 + 1$$ --- ### Part b) 7. Compare each term of sequence b) with $n^2$: - For $n=1$: $n^2=1$, sequence b) term = 2 - For $n=2$: $n^2=4$, sequence b) term = 8 - For $n=3$: $n^2=9$, sequence b) term = 18 - For $n=4$: $n^2=16$, sequence b) term = 32 8. Check the ratio of sequence b) terms to $n^2$: - $\frac{2}{1} = 2$ - $\frac{8}{4} = 2$ - $\frac{18}{9} = 2$ - $\frac{32}{16} = 2$ 9. Each term in sequence b) is twice $n^2$. 10. Therefore, the nth term for sequence b) is: $$b_n = 2n^2$$ --- **Final answers:** - a) $a_n = n^2 + 1$ - b) $b_n = 2n^2$