To solve the given problem, we start by analyzing the sequence $\{a_n\}$ defined by:

$a_n = 2 - \frac{1}{n^2 + \sqrt{n^4 + \frac{1}{4}}}$

Step 1: Simplify the expression for $a_n$

First, observe the term inside the sequence:

$b_n = \frac{1}{n^2 + \sqrt{n^4 + \frac{1}{4}}}$

Let's simplify $b_n$. The denominator can be rewritten to simplify using rationalization:

$n^2 + \sqrt{n^4 + \frac{1}{4}} = \frac{(n^2 + \sqrt{n^4 + \frac{1}{4}})(n^2 - \sqrt{n^4 + \frac{1}{4}})}{n^2 - \sqrt{n^4 + \frac{1}{4}}}$

Using the identity $(a+b)(a-b) = a^2 - b^2$, we find:

= \frac{-\frac{1}{4}}{n^2 - \sqrt{n^4 + \frac{1}{4}}}$$ Thus, the denominator is effectively: $$n^2 + \sqrt{n^4 + \frac{1}{4}} = \frac{n^4 + \frac{1}{4}}{n^2 - \sqrt{n^4 + \frac{1}{4}}} \approx 2n^2 \quad \text{(using approximation)}$$ $$b_n \approx \frac{1}{2n^2}$$ ### Step 2: Approximating $$a_n$$ Using the above approximation for $$b_n$$: $$a_n \approx 2 - b_n \approx 2 - \frac{1}{2n^2}$$ ### Step 3: Finding the sum $$\frac{1}{2} \left( \sqrt{a_1} + \sqrt{a_2} + \ldots + \sqrt{a_{119}} \right)$$ Given the expression: $$\sqrt{a_n} \approx \sqrt{2 - \frac{1}{2n^2}}$$ For large $$n$$, $$\sqrt{2 - \frac{1}{2n^2}} \approx \sqrt{2}$$ (since $$\frac{1}{2n^2}$$ is very small). Now, we approximate: $$\frac{1}{2} \left( \sqrt{a_1} + \sqrt{a_2} + \ldots + \sqrt{a_{119}} \right) \approx \frac{1}{2} \left( 119 \times \sqrt{2} \right) = \frac{119 \sqrt{2}}{2}$$ $$= 59.5 \sqrt{2}$$ ### Final Answer $$\frac{1}{2} \left( \sqrt{a_1} + \sqrt{a_2} + \ldots + \sqrt{a_{119}} \right) \approx 59.5 \sqrt{2}$$ ### Additional Questions and Tips 1. How does rationalizing the denominator help in simplifying expressions? 2. Why is it useful to approximate sequences when $$n$$ is large? 3. What are other methods to approximate square roots of expressions like $$\sqrt{2 - \frac{1}{2n^2}}$$? 4. Can we find a general expression for $$\sqrt{a_n}$$ without approximation? 5. How do small terms affect the accuracy of approximations in series or sequences? **Tip**: Always check the behavior of sequences as $$n \to \infty$$ to see if terms approach a limit, simplifying calculations.