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

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

Aicharya · Accepted Answer

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

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function satisfying $f(0)=1$ and $f(2…

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