To find the first three terms and the formula for the sequence $a_n = \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right)$, let's break this down.

Step 1: Simplifying each term

Each factor in the product takes the form: $1 - \frac{1}{k^2} = \frac{k^2 - 1}{k^2} = \frac{(k-1)(k+1)}{k^2}$ for each $k$ from 2 to $n$.

Step 2: Writing the product for $a_n$

So, the sequence $a_n$ can be written as: $a_n = \prod_{k=2}^{n} \left( 1 - \frac{1}{k^2} \right) = \prod_{k=2}^{n} \frac{(k-1)(k+1)}{k^2}$

Step 3: Calculating the first three terms

1. For $n = 2$: $a_2 = \left(1 - \frac{1}{4}\right) = \frac{3}{4}$

2. For $n = 3$: $a_3 = \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) = \frac{3}{4} \times \frac{8}{9} = \frac{24}{36} = \frac{2}{3}$

3. For $n = 4$: $a_4 = \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right) = \frac{3}{4} \times \frac{8}{9} \times \frac{15}{16} = \frac{360}{576} = \frac{5}{8}$

Step 4: General Formula for $a_n$

After simplification, the sequence $a_n$ can be expressed as: $a_n = \frac{2}{n+1}$

Summary:

• First three terms: $a_2 = \frac{3}{4}, \quad a_3 = \frac{2}{3}, \quad a_4 = \frac{5}{8}$
• General formula: $a_n = \frac{2}{n+1}$

Would you like more details or have any further questions?

Here are five related questions:

1. How would the formula for $a_n$ change if the product started from $k=1$ instead of $k=2$?
2. Can you find the limit of $a_n$ as $n \to \infty$?
3. What is the sum of the sequence $a_n$ from $n=2$ to infinity?
4. How would the sequence change if each term was raised to a power?
5. Can the sequence be expressed as a continued fraction?

Tip: Recognizing patterns in product sequences often leads to telescoping series, which simplify the process of finding general formulas.