To solve this problem, let's break it down step-by-step.

Step 1: Expand $(1 + 2x)^{10}$ up to the fourth term in powers of $x$.

This is a binomial expansion problem where we expand $(1 + 2x)^{10}$ using the Binomial Theorem: $(1 + 2x)^{10} = \sum_{k=0}^{10} \binom{10}{k} \cdot (1)^{10-k} \cdot (2x)^k$

We are asked to expand up to and including the fourth term. So, we need the terms for $k = 0, 1, 2, 3,$ and $4$.

Terms Calculation:

1. For $k = 0$: $\binom{10}{0} \cdot (1)^{10} \cdot (2x)^0 = 1$

2. For $k = 1$: $\binom{10}{1} \cdot (1)^{9} \cdot (2x)^1 = 10 \cdot 2x = 20x$

3. For $k = 2$: $\binom{10}{2} \cdot (1)^{8} \cdot (2x)^2 = 45 \cdot 4x^2 = 180x^2$

4. For $k = 3$: $\binom{10}{3} \cdot (1)^{7} \cdot (2x)^3 = 120 \cdot 8x^3 = 960x^3$

5. For $k = 4$: $\binom{10}{4} \cdot (1)^{6} \cdot (2x)^4 = 210 \cdot 16x^4 = 3360x^4$

Thus, the expansion of $(1 + 2x)^{10}$ up to the fourth term is: $1 + 20x + 180x^2 + 960x^3 + 3360x^4$

Step 2: Use the expansion to evaluate $(1.0002)^{10}$

To find $(1.0002)^{10}$, we let $x = 0.0001$ in our expansion $(1 + 2x)^{10}$.

Substitute $x = 0.0001$ into the expression: $1 + 20(0.0001) + 180(0.0001)^2 + 960(0.0001)^3 + 3360(0.0001)^4$

Now, calculate each term:

1. $20 \times 0.0001 = 0.002$
2. $180 \times (0.0001)^2 = 180 \times 0.00000001 = 0.0000018$
3. $960 \times (0.0001)^3 = 960 \times 0.000000000001 = 0.000000096$
4. $3360 \times (0.0001)^4 = 3360 \times 0.0000000000000001 = 0.000000000336$

Add these terms together: $1 + 0.002 + 0.0000018 + 0.000000096 + 0.000000000336 \approx 1.0020019$

Final Answer:

To four decimal places, $(1.0002)^{10} \approx 1.0020$.

Would you like a detailed explanation of any specific part? Here are some related questions that might expand your understanding:

1. How does the Binomial Theorem work in general for any expansion $(a + b)^n$?
2. What are binomial coefficients, and how are they calculated?
3. Can the expansion be used for other values of $n$ or different terms in the series?
4. What are some real-world applications of binomial expansion?
5. Why do we approximate $(1.0002)^{10}$ using the first few terms only?

Tip: When dealing with binomial expansions for small changes (like $1.0002$), the first few terms usually provide a very close approximation.