Let f: $\mathbb{R} \to \mathbb{R}$ be a continuous function satisfying f(0) = 1 and f(2x) - f(x) = x for all x $\in \mathbb{R}$. If $\lim_{n\to\infty} [f(x) - f(x/2^n)] = G(x)$ then $\sum_{r=1}^{10} G(r^2)$ is equal to:

Step 1: Express f(x) in terms of f(x/ $2^n$)

We are given the functional equation $f(2x) - f(x) = x$. We can rewrite this as $f(y) - f(y/2) = y/2$ by substituting $y = 2x$. Using this, we can express the difference $f(x) - f(x/2^n)$ as a telescoping sum.

Step 2: Substitute the functional equation into the sum

For each term in the sum, we apply the relation $f(y) - f(y/2) = y/2$. Here, $y = x/2^{k-1}$, so $y/2 = x/2^k$. Thus, each term $f(x/2^{k-1}) - f(x/2^k)$ becomes $x/2^k$.

Step 3: Evaluate the sum

The sum $\sum_{k=1}^{n} (1/2)^k$ is a geometric series with first term $a = 1/2$ and common ratio $r = 1/2$. The sum of the first $n$ terms of a geometric series is given by $a(1-r^n)/(1-r)$.

Step 4: Find G(x) by taking the limit

As $n \to \infty$, the term $(1/2)^n$ approaches 0. Therefore, $G(x)$ simplifies to $x$. The continuity of $f$ and $f(0)=1$ are consistent with this result, as $f(x/2^n) \to f(0) = 1$ as $n \to \infty$.

Step 5: Calculate the final sum

Since $G(x) = x$, we have $G(r^2) = r^2$. We need to calculate the sum of the first 10 squares. The formula for the sum of the first $N$ squares is $N(N+1)(2N+1)/6$.

Step 6: Apply the sum of squares formula

Substitute $N=10$ into the formula for the sum of the first $N$ squares to get the final result.

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