Let f: mathbbR → mathbbR be a continuous function satisfying f(0)=1 and f(2 x)-f( x)= x for all x ∈ mathbbR. If lim _n → ∞ f(x)-f ( x 2^n ) =G(x),…

$\begin{aligned}
& f(2 x)-f(x)=x \
& f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2} \
& \mathrm{f}\left(\frac{\mathrm{x}}{2}\right)-\mathrm{f}\left(\frac{\mathrm{x}}{4}\right)=\frac{\mathrm{x}}{4} \
& \mathrm{f}\left(\frac{\mathrm{x}}{4}\right)-\mathrm{f}\left(\frac{\mathrm{x}}{8}\right)=\frac{\mathrm{x}}{8} \
& f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n} \
& f(2 x)-f\left(\frac{x}{2^n}\right)=x\left{\frac{1-\left(\frac{1}{2}\right)^{n-1}}{1-\frac{1}{2}}\right} \
& f(x)-f\left(\frac{x}{2^n}\right)=2 x\left(1-\left(\frac{1}{2}\right)^{n+1}\right) \
& \mathrm{f}(\mathrm{x})+\mathrm{x}-\mathrm{f}\left(\frac{\mathrm{x}}{2^{\mathrm{n}}}\right)=2 \mathrm{x}\left(1-\left(\frac{1}{2}\right)^{\mathrm{n}+1}\right) \
& \lim {n \rightarrow \infty}\left(f(x)-f\left(\frac{x}{2^n}\right)\right)=\lim {n \rightarrow \infty}\left(2 x\left(1-\left(\frac{1}{2}\right)^{n+1}\right)-x\right) \
& G(x)=x \
& \sum{r=1}^{10} G\left(r^2\right)=\sum{r=1}^{10} r^2=385
\end{aligned}$