Find the first three non-zero terms of the Taylor series about x = 0 for the following function

f(x)=〖(1+x)〗^100/(〖(1-2x)〗^40 〖(1+2x)〗^60 )
Show your working. Use the Taylor series (1 + x)^α 

We are tasked with finding the first three non-zero terms of the Taylor series about \(x = 0\) for the function:

\[
f(x) = \frac{(1 + x)^{100}}{(1 - 2x)^{40} (1 + 2x)^{60}}.
\]

To do this, we will expand each factor in the numerator and denominator using the binomial series. Specifically, we will use the Taylor series expansion of \((1 + x)^\alpha\) around \(x = 0\), which is:

\[
(1 + x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2}x^2 + \cdots.
\]

We will now expand each component of the given function step-by-step.

### Step 1: Expansion of \((1 + x)^{100}\)
The Taylor series expansion of \((1 + x)^{100}\) around \(x = 0\) is:

\[
(1 + x)^{100} = 1 + 100x + \frac{100(100 - 1)}{2}x^2 + \cdots.
\]

Simplifying the quadratic term:

\[
(1 + x)^{100} = 1 + 100x + 4950x^2 + \cdots.
\]

### Step 2: Expansion of \((1 - 2x)^{40}\)
Now, expand \((1 - 2x)^{40}\) using the binomial series:

\[
(1 - 2x)^{40} = 1 - 40(2x) + \frac{40(40 - 1)}{2}(2x)^2 + \cdots.
\]

Simplifying:

\[
(1 - 2x)^{40} = 1 - 80x + \frac{40 \times 39}{2} \times 4x^2 + \cdots = 1 - 80x + 3120x^2 + \cdots.
\]

### Step 3: Expansion of \((1 + 2x)^{60}\)
Similarly, expand \((1 + 2x)^{60}\) using the binomial series:

\[
(1 + 2x)^{60} = 1 + 60(2x) + \frac{60(60 - 1)}{2}(2x)^2 + \cdots.
\]

Simplifying:

\[
(1 + 2x)^{60} = 1 + 120x + \frac{60 \times 59}{2} \times 4x^2 + \cdots = 1 + 120x + 7080x^2 + \cdots.
\]

### Step 4: Combine the expansions
Now, combine the expansions of the numerator and denominator. The function becomes:

\[
f(x) = \frac{(1 + 100x + 4950x^2 + \cdots)}{(1 - 80x + 3120x^2 + \cdots)(1 + 120x + 7080x^2 + \cdots)}.
\]

To simplify this, let's first expand the denominator up to the quadratic terms. First, multiply the two expansions:

\[
(1 - 80x + 3120x^2)(1 + 120x + 7080x^2).
\]

Using distributive property, we get:

\[
1 + 120x - 80x + 7080x^2 - 9600x^2 + 3120x^2 + \cdots = 1 + 40x + 3600x^2 + \cdots.
\]

Thus, the denominator simplifies to:

\[
1 + 40x + 3600x^2 + \cdots.
\]

### Step 5: Divide the numerator by the denominator
Now, divide the expansion of the numerator by the expansion of the denominator, using long division:

\[
f(x) = \frac{1 + 100x + 4950x^2 + \cdots}{1 + 40x + 3600x^2 + \cdots}.
\]

Performing the division term by term:

- The constant term is \(1\).
- The linear term is \((100x - 40x) = 60x\).
- The quadratic term is \((4950x^2 - 40 \times 60x^2 - 3600x^2) = 4950x^2 - 2400x^2 - 3600x^2 = -50x^2\).

Thus, the first three non-zero terms of the Taylor series are:

\[
f(x) = 1 + 60x - 50x^2 + \cdots.
\]

### Final Answer:
The first three non-zero terms of the Taylor series are:

\[
f(x) = 1 + 60x - 50x^2 + \cdots.
\]

Would you like more details or have any questions?

---

Here are 5 related questions to explore:

1. What is the general form of the binomial expansion for \((1 + x)^\alpha\)?
2. How do we handle higher-order terms in Taylor series expansions?
3. What are the applications of Taylor series in approximating functions?
4. How does the Taylor series change if we expand around a point \(x = a\) instead of \(x = 0\)?
5. How would the expansion change if the powers in the denominator were different?

**Tip:** When expanding functions, it's often easier to expand components separately before combining, as shown in this example.

Find the first three non-zero terms of the Taylor series about x = 0 for the following function: f(x) = (1 + x)^100 / [(1 - 2x)^40 (1 + 2x)^60]. Show your working. Use the Taylor series (1 + x)^α.

Explore the first three non-zero terms of the Taylor series for the function f(x) = (1 + x)^100 / [(1 - 2x)^40 (1 + 2x)^60]. This detailed solution utilizes binomial expansion and Taylor series, covering advanced calculus concepts.

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