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(B) Given $f(2x) - f(x) = x$. Replacing $x$ with $\frac{x}{2}, \frac{x}{4}, \dots, \frac{x}{2^n}$,we get:$f(x) - f(\frac{x}{2}) = \frac{x}{2}$$f(\frac

Vedclass · Accepted Answer

$385$

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(B) Given $f(2x) - f(x) = x$. Replacing $x$ with $\frac{x}{2}, \frac{x}{4}, \dots, \frac{x}{2^n}$,we get:$f(x) - f(\frac{x}{2}) = \frac{x}{2}$$f(\frac{x}{2}) - f(\frac{x}{4}) = \frac{x}{4}$$f(\frac{x}{4}) - f(\frac{x}{8}) = \frac{x}{8}$$f(\frac{x}{2^{n-1}}) - f(\frac{x}{2^n}) = \frac{x}{2^n}$Summing these equations,we get:$f(x) - f(\frac{x}{2^n}) = \sum_{k=1}^{n} \frac{x}{2^k} = x \left( \frac{1/2(1 - (1/2)^n)}{1 - 1/2} \right) = x(1 - (\frac{1}{2})^n)$.Taking the limit as $n \rightarrow \infty$:$G(x) = \lim_{n \rightarrow \infty} (f(x) - f(\frac{x}{2^n})) = x(1 - 0) = x$.Thus,$\sum_{r=1}^{10} G(r^2) = \sum_{r=1}^{10} r^2 = \frac{10(11)(21)}{6} = 385$.

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

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