Key Concepts and Formulas

• Telescoping Sum: A series where intermediate terms cancel out, simplifying the sum.
• Geometric Series: The sum of a finite or infinite geometric series.
• Continuity of Functions: A function is continuous at a point if the limit exists at that point and equals the function's value.

Step-by-Step Solution

Step 1: Understand the given functional equation and initial condition. We are given a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ with $f(0)=1$ and the functional equation $f(2x) - f(x) = x$ for all $x \in \mathbb{R}$.

Step 2: Manipulate the functional equation to reveal a pattern. The given equation is $f(2x) - f(x) = x$. To work with the limit expression $f(x) - f(\frac{x}{2^n})$, we need to express the difference in a form that relates $f(x)$ to $f(x/2)$. Let's replace $x$ with $x/2$ in the given functional equation: $f(2(x/2)) - f(x/2) = x/2$ $f(x) - f(x/2) = x/2$ ..... (1)

Step 3: Apply the manipulated equation iteratively. Now, we can apply this new form of the equation repeatedly to get a series. Replace $x$ with $x/2$ in equation (1): $f(x/2) - f((x/2)/2) = (x/2)/2$ $f(x/2) - f(x/4) = x/4 = x/2^2$ ..... (2)

Replace $x$ with $x/4$ in equation (1): $f(x/4) - f((x/4)/2) = (x/4)/2$ $f(x/4) - f(x/8) = x/8 = x/2^3$ ..... (3)

We can see a pattern emerging. For any positive integer $k$, replacing $x$ with $x/2^{k-1}$ in equation (1) gives: $f(x/2^{k-1}) - f(x/2^k) = (x/2^{k-1})/2 = x/2^k$

Step 4: Form a telescoping sum. We want to find the limit of $f(x) - f(\frac{x}{2^n})$. Let's consider the sum of the first $n$ terms of the pattern we found. We have: $f(x) - f(x/2) = x/2$ $f(x/2) - f(x/4) = x/4$ $f(x/4) - f(x/8) = x/8$ ... $f(x/2^{n-1}) - f(x/2^n) = x/2^n$

Summing these $n$ equations: $(f(x) - f(x/2)) + (f(x/2) - f(x/4)) + \ldots + (f(x/2^{n-1}) - f(x/2^n)) = x/2 + x/4 + \ldots + x/2^n$

The intermediate terms on the left-hand side cancel out due to the telescoping nature of the sum. This leaves us with: $f(x) - f(x/2^n) = \frac{x}{2} + \frac{x}{2^2} + \frac{x}{2^3} + \ldots + \frac{x}{2^n}$

Step 5: Evaluate the limit to find G(x). We are given $G(x) = \lim_{n \rightarrow \infty} \{f(x) - f(\frac{x}{2^n})\}$. Using the result from Step 4: $G(x) = \lim_{n \rightarrow \infty} \left( \frac{x}{2} + \frac{x}{2^2} + \frac{x}{2^3} + \ldots + \frac{x}{2^n} \right)$

The expression inside the limit is a finite geometric series with first term $a = x/2$, common ratio $r = 1/2$, and $n$ terms. The sum of a finite geometric series is $S_n = a \frac{1-r^n}{1-r}$. So, $\frac{x}{2} + \frac{x}{2^2} + \ldots + \frac{x}{2^n} = \frac{x}{2} \frac{1 - (1/2)^n}{1 - 1/2} = \frac{x}{2} \frac{1 - (1/2)^n}{1/2} = x(1 - (1/2)^n)$.

Now, we take the limit as $n \rightarrow \infty$: $G(x) = \lim_{n \rightarrow \infty} x(1 - (1/2)^n)$ Since $\lim_{n \rightarrow \infty} (1/2)^n = 0$, we get: $G(x) = x(1 - 0) = x$

Let's verify this result using the continuity of $f$. From $f(x) - f(x/2^n) = x(1 - (1/2)^n)$, as $n \rightarrow \infty$, $x/2^n \rightarrow 0$. Since $f$ is continuous, $\lim_{n \rightarrow \infty} f(x/2^n) = f(\lim_{n \rightarrow \infty} x/2^n) = f(0)$. We are given $f(0)=1$. So, $G(x) = \lim_{n \rightarrow \infty} \{f(x) - f(\frac{x}{2^n})\} = f(x) - \lim_{n \rightarrow \infty} f(\frac{x}{2^n}) = f(x) - f(0) = f(x) - 1$. Therefore, $G(x) = f(x) - 1$. From our earlier derivation, $G(x) = x$. This implies $f(x) - 1 = x$, so $f(x) = x+1$. Let's check if $f(x)=x+1$ satisfies the original conditions: $f(0) = 0+1 = 1$ (Satisfied) $f(2x) - f(x) = (2x+1) - (x+1) = 2x+1-x-1 = x$ (Satisfied) So, $f(x)=x+1$ is indeed the function. And $G(x) = f(x) - f(0) = (x+1) - 1 = x$.

Step 6: Calculate G(r^2) for a general r. We found that $G(x) = x$. Therefore, $G(r^2) = r^2$.

Step 7: Compute the sum $\sum_{r=1}^{10} G(r^2)$. We need to calculate $\sum_{r=1}^{10} G(r^2)$. Substituting $G(r^2) = r^2$: $\sum_{r=1}^{10} G(r^2) = \sum_{r=1}^{10} r^2$

This is the sum of the squares of the first 10 natural numbers. The formula for the sum of the first $N$ squares is $\sum_{r=1}^{N} r^2 = \frac{N(N+1)(2N+1)}{6}$. Here, $N=10$. $\sum_{r=1}^{10} r^2 = \frac{10(10+1)(2 \cdot 10+1)}{6}$ $\sum_{r=1}^{10} r^2 = \frac{10(11)(21)}{6}$ $\sum_{r=1}^{10} r^2 = \frac{2310}{6}$ $\sum_{r=1}^{10} r^2 = 385$

Common Mistakes & Tips

• Incorrectly forming the geometric series: Ensure the first term and common ratio are correctly identified after manipulating the functional equation.
• Forgetting the limit process: The definition of $G(x)$ involves a limit. Make sure to evaluate this limit correctly, especially when dealing with terms that tend to zero.
• Algebraic errors in summation formula: Double-check the application of the sum of squares formula to avoid calculation mistakes.

Summary

The problem involves a functional equation and a limit. We first manipulated the functional equation $f(2x) - f(x) = x$ by substituting $x/2$ for $x$ to obtain $f(x) - f(x/2) = x/2$. By iteratively applying this relation and summing the resulting terms, we constructed a telescoping series for $f(x) - f(x/2^n)$. Taking the limit as $n \rightarrow \infty$, we found that $G(x) = x$. Consequently, $G(r^2) = r^2$. The final step was to compute the sum of the squares of the first 10 natural numbers using the standard formula, which yielded 385.

The final answer is \boxed{385} which corresponds to option (C).