(i) First Three Terms in the Binomial Expansion of $(1 - x)^{-2}$

The binomial expansion of $(1 - x)^{-k}$ is given by: $(1 - x)^{-k} = \sum_{n=0}^{\infty} \binom{k+n-1}{n} x^n$ For $(1 - x)^{-2}$, we substitute $k = 2$: $(1 - x)^{-2} = \sum_{n=0}^{\infty} \binom{n+1}{n} x^n$ The binomial coefficients $\binom{n+1}{n}$ simplify as follows: $\binom{n+1}{n} = n+1$ Thus, the first three terms of the expansion in ascending powers of $x$ are: $(1 - x)^{-2} = 1 + 2x + 3x^2 + \cdots$

(ii) Coefficient of $x^n$

From the general binomial expansion, the coefficient of $x^n$ in the expansion of $(1 - x)^{-2}$ is given by: $\binom{n+1}{n} = n+1$ Thus, the coefficient of $x^n$ is $n+1$.

(iii) Range of Values of $x$ for the Expansion to be Valid

The binomial expansion $(1 - x)^{-k}$ is valid for: $|x| < 1$ Therefore, for $(1 - x)^{-2}$, the range of $x$ is: $|x| < 1$

(iv) Finding the Value of $\sum_{n=1}^{\infty} \frac{n}{2^n}$

We are asked to find: $\sum_{n=1}^{\infty} \frac{n}{2^n}$ This is a standard summation. To solve it, we can express it as a power series derivative.

Start with the geometric series: $\sum_{n=0}^{\infty} x^n = \frac{1}{1 - x} \quad \text{for} \ |x| < 1$ Now differentiate both sides with respect to $x$: $\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1 - x)^2}$ Multiply by $x$ to adjust the powers: $\sum_{n=1}^{\infty} n x^n = \frac{x}{(1 - x)^2}$ Now substitute $x = \frac{1}{2}$ to find the value of the given series: $\sum_{n=1}^{\infty} \frac{n}{2^n} = \frac{\frac{1}{2}}{\left(1 - \frac{1}{2}\right)^2} = \frac{\frac{1}{2}}{\left(\frac{1}{2}\right)^2} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2$

Thus, the value of the sum is: $\sum_{n=1}^{\infty} \frac{n}{2^n} = 2$

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Would you like further details on any part of this explanation or have any questions?

Additional Questions to Explore:

1. How do we derive the general form of binomial expansion for negative exponents?
2. What is the binomial coefficient $\binom{k+n-1}{n}$ in more detail?
3. How do we extend the expansion of $(1 - x)^{-2}$ beyond the first three terms?
4. Can you explore the radius of convergence for similar binomial expansions?
5. How can we apply these expansions to approximate functions?

Tip: When working with series and summations, geometric series and their derivatives are key tools for simplifying complex sums.