We are asked to find a formula for the $n$th term of two sequences given in the image.

(a) Sequence: $\frac{1}{1}, -\frac{1}{4}, \frac{1}{9}, \dots$

This sequence alternates in sign, and the denominators are perfect squares: $\frac{1}{1^2}, -\frac{1}{2^2}, \frac{1}{3^2}, \dots$

Thus, we can describe the $n$-th term $a_n$ as: $a_n = (-1)^{n+1} \cdot \frac{1}{n^2}$ The $(-1)^{n+1}$ introduces the alternating sign, and $\frac{1}{n^2}$ gives the denominator as a square.

(b) Sequence: $\frac{4}{7}, \frac{5}{8}, \frac{6}{9}, \dots$

For this sequence, we notice that both the numerators and the denominators increase by 1. Specifically, for the $n$-th term:

• The numerator starts at 4 and increases by 1, so it is $3 + n$.
• The denominator starts at 7 and increases by 1, so it is $6 + n$.

Thus, the $n$-th term $b_n$ is: $b_n = \frac{3+n}{6+n}$

Final Formulas:

• $a_n = (-1)^{n+1} \cdot \frac{1}{n^2}$
• $b_n = \frac{3+n}{6+n}$

Would you like any further clarification or details on these solutions?

Here are 5 related questions to further your understanding:

1. How do you identify patterns in sequences for numerators and denominators separately?
2. How would the formula change if the signs in the first sequence did not alternate?
3. Can you generalize the formula for sequences with cubic denominators?
4. How would you find the formula if the sequence had alternating signs but an exponential denominator?
5. What approach would you take to find the formula for non-fraction sequences?

Tip: Always look for patterns in both the numerator and the denominator when dealing with fractional sequences.