(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}$$

Again, Replace $x \rightarrow \frac{x}{2}$

$f\left(\frac{x}{2}\right)-f\left(\frac{x}{2^2}\right)=\frac{x}{2^2}\quad\text{...... (2)}$

: $\qquad$ : $\qquad$ :

$$\begin{aligned} & f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n} \\ & \text { Adding }(1)+(2)+(3)+\ldots+(\mathrm{n}) \\ & \text { We get } f(x)-f\left(\frac{x}{2^n}\right)=\frac{x}{2}+\frac{x}{2^2}+\ldots .+\frac{x}{2^n} \\ & \lim _{n \rightarrow \infty}\left(f(x)-f\left(\frac{x}{2^n}\right)\right)=\lim _{n \rightarrow \infty}\left(\frac{x}{2}+\frac{x}{2^n}+\ldots+\frac{x}{2^n}\right) \end{aligned}$$

$$\begin{aligned} & f(x)-f(0)=\frac{\frac{x}{2}}{\frac{1}{2}} \\ & \Rightarrow G(x)=x \\ & \Rightarrow G\left(r^2\right)=r^2 \\ & \Rightarrow \sum_{r=1}^{10} G\left(r^2\right)=\sum_{r=1}^{10} G\left(r^2\right) \\ & =\frac{(10)(11)(21)}{6}=(55) 7 \\ & \Rightarrow 385 \end{aligned}$$

19:T50e,

$$\begin{aligned} & f(0)=1, f(2 x)-f(x)=x \\ & \text { प्रतिस्थापित करें } x \rightarrow \frac{x}{2} \\ & f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2}\quad\text{..... (1)} \end{aligned}$$

पुनः, प्रतिस्थापित करें $x \rightarrow \frac{x}{2}$

$f\left(\frac{x}{2}\right)-f\left(\frac{x}{2^2}\right)=\frac{x}{2^2}\quad\text{...... (2)}$

: $\qquad$ : $\qquad$ :

$$\begin{aligned} & f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n} \\ & \text { जोड़ने पर }(1)+(2)+(3)+\ldots+(\mathrm{n}) \\ & \text { हमें मिलता है } 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

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