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

We are given a continuous function $$f:\mathbb{R}\to\mathbb{R}$$ that satisfies
$$f(2x)-f(x)=x \quad\text{for all }x\in\mathbb{R}, \qquad f(0)=1.$$(1)

The limit to be studied is
$$G(x)=\lim_{n\to\infty}\Bigl\{f(x)-f\!\left(\dfrac{x}{2^{\,n}}\right)\Bigr\}.$$(2)

Case 1: Find a closed form for $$f(x)-f\!\left(\dfrac{x}{2^{\,n}}\right)\,$$ when $$n$$ is a positive integer.

Set $$t=\dfrac{x}{2^{\,k}}$$ in the functional equation $$(1).$$ Then $$f(2t)-f(t)=t$$ becomes
$$f\!\left(\dfrac{x}{2^{\,k-1}}\right)-f\!\left(\dfrac{x}{2^{\,k}}\right)=\dfrac{x}{2^{\,k}} \quad (k\ge 1).$$(3)

Add equations $$(3)$$ for $$k=1,2,\dots ,n$$:

$$\bigl[f(x)-f\!\left(\tfrac{x}{2}\right)\bigr]+\bigl[f\!\left(\tfrac{x}{2}\right)-f\!\left(\tfrac{x}{2^{\,2}}\right)\bigr]+\cdots+\bigl[f\!\left(\tfrac{x}{2^{\,n-1}}\right)-f\!\left(\tfrac{x}{2^{\,n}}\right)\bigr]$$ $$\;=\;\dfrac{x}{2^{\,1}}+\dfrac{x}{2^{\,2}}+\cdots+\dfrac{x}{2^{\,n}}.$$(4)

The left side telescopes, leaving

$$f(x)-f\!\left(\dfrac{x}{2^{\,n}}\right)=x\!\left(\dfrac{1}{2}+\dfrac{1}{2^{\,2}}+\cdots+\dfrac{1}{2^{\,n}}\right).$$(5)

The geometric-series sum is

$$\dfrac{1}{2}+\dfrac{1}{2^{\,2}}+\cdots+\dfrac{1}{2^{\,n}}=1-\dfrac{1}{2^{\,n}}.$$(6)

Substituting $$(6)$$ into $$(5)$$ gives the exact identity

$$f(x)-f\!\left(\dfrac{x}{2^{\,n}}\right)=x\Bigl(1-\dfrac{1}{2^{\,n}}\Bigr).$$(7)

Case 2: Pass to the limit $$n\to\infty$$ in $$(7).$$

Because $$\dfrac{1}{2^{\,n}}\to 0,$$ we get

$$\lim_{n\to\infty}\Bigl\{f(x)-f\!\left(\dfrac{x}{2^{\,n}}\right)\Bigr\}=x.$$(8)

Hence, from definition $$(2),$$

$$G(x)=x \quad\text{for every }x\in\mathbb{R}.$$(9)

Case 3: Evaluate the required sum.

Using $$(9),$$

$$\sum_{r=1}^{10} G(r^{2})=\sum_{r=1}^{10} r^{2}.$$(10)

The formula for the sum of the first $$n$$ squares is
$$\sum_{r=1}^{n} r^{2}=\dfrac{n(n+1)(2n+1)}{6}.$$(11)

With $$n=10,$$ equation $$(11)$$ gives

$$\sum_{r=1}^{10} r^{2}=\dfrac{10\cdot11\cdot21}{6}=385.$$(12)

Therefore,
$$\sum_{r=1}^{10} G(r^{2})=385.$$(13)

The correct option is Option B (385).