The expression in the image is:

$\sqrt{1 + \left(\frac{x}{3}\right)^2}$

We can expand this expression using the binomial theorem. To do so, first rewrite it in a form suitable for expansion:

$\sqrt{1 + \left(\frac{x}{3}\right)^2} = \left(1 + \left(\frac{x}{3}\right)^2\right)^{\frac{1}{2}}$

The binomial expansion of $(1 + u)^n$ where $|u| < 1$ and $n$ is any real number, is given by:

$(1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \dots$

For our expression:

• $n = \frac{1}{2}$
• $u = \left(\frac{x}{3}\right)^2$

Now, expand the first three terms:

$\left(1 + \left(\frac{x}{3}\right)^2\right)^{\frac{1}{2}} \approx 1 + \frac{1}{2}\left(\frac{x}{3}\right)^2 + \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right)}{2} \left(\frac{x}{3}\right)^4$

Let's simplify these terms:

1. The first term is $1$.
2. The second term is $\frac{1}{2}\left(\frac{x}{3}\right)^2 = \frac{x^2}{18}$.
3. The third term involves calculating $\frac{\frac{1}{2} \left(\frac{-1}{2}\right)}{2} = -\frac{1}{8}$, so the term becomes:

$-\frac{1}{8} \left(\frac{x^2}{9}\right)^2 = -\frac{x^4}{648}$

So, the first three terms of the binomial expansion are:

$1 + \frac{x^2}{18} - \frac{x^4}{648}$

Would you like more details or have any further questions?

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Here are five related questions you might find interesting:

1. How does the binomial theorem apply to non-integer exponents?
2. What conditions must be satisfied for the binomial expansion to converge?
3. How would the expansion differ if the exponent was an integer?
4. Can you use the binomial expansion to approximate the value of functions like $\sqrt{1+x}$ for small $x$?
5. How does the number of terms affect the accuracy of the binomial approximation?

Tip: When using the binomial expansion for small perturbations (e.g., small $x$), you can often achieve accurate approximations by retaining only the first few terms of the series.