Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function satisfying $f(0)=1$ and $f(2 x)-f(x)=x$ for all $x \in \mathbb{R}$. If $\lim _\limits{n \rightarrow \infty}\left\{f(x)-f\left(\frac{x}{2^n}\right)\right\}=G(x)$, then $\sum_\limits{r=1}^{10} G\left(r^2\right)$ is equal to

<p>$$\begin{aligned} & f(0)=1, f(2 x)-f(x)=x \\ & \text { Replace } x \rightarrow \frac{x}{2} \\ & f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2}\quad\text{..... (1)} \end{aligned}$$</p> <p>Again, Replace $x \rightarrow \frac{x}{2}$</p> <p>$f\left(\frac{x}{2}\right)-f\left(\frac{x}{2^2}\right)=\frac{x}{2^2}\quad\text{...... (2)}$</p> <p>: $\qquad$ : $\qquad$ :</p> <p>: $\qquad$ : $\qquad$ :</p> <p>: $\qquad$ : $\qquad$ :</p> <p>$$\begin{aligned} & f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n} \\ & \text { Adding }(1)+(2)+(3)+\ldots+(\mathrm{n}) \\ & \text { We get } f(x)-f\left(\frac{x}{2^n}\right)=\frac{x}{2}+\frac{x}{2^2}+\ldots .+\frac{x}{2^n} \\ & \lim _{n \rightarrow \infty}\left(f(x)-f\left(\frac{x}{2^n}\right)\right)=\lim _{n \rightarrow \infty}\left(\frac{x}{2}+\frac{x}{2^n}+\ldots+\frac{x}{2^n}\right) \end{aligned}$$</p> <p>$$\begin{aligned} & f(x)-f(0)=\frac{\frac{x}{2}}{\frac{1}{2}} \\ & \Rightarrow G(x)=x \\ & \Rightarrow G\left(r^2\right)=r^2 \\ & \Rightarrow \sum_{r=1}^{10} G\left(r^2\right)=\sum_{r=1}^{10} G\left(r^2\right) \\ & =\frac{(10)(11)(21)}{6}=(55) 7 \\ & \Rightarrow 385 \end{aligned}$$</p>