Question

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 $\lim_{n \to \infty} \left\{ f(x) - f\left( \frac{x}{2^n} \right) \right\} = G(x)$, then $\sum_{r=1}^{10} G(r^2)$ is equal to

• A. 540
• B. 385
• C. 420
• D. 215

Hint:

Use the sum of squares formula to calculate the sum.

Answer 1

The Correct Option is B

1. Use the given functional equation: \[ f(2x) - f(x) = x \]
2. Express $f(x)$ in terms of $f\left( \frac{x}{2^n} \right)$: \[ f(x) - f\left( \frac{x}{2} \right) = \frac{x}{2} \] \[ f\left( \frac{x}{2} \right) - f\left( \frac{x}{4} \right) = \frac{x}{4} \] \[ f\left( \frac{x}{4} \right) - f\left( \frac{x}{8} \right) = \frac{x}{8} \] \[ \vdots \] \[ f\left( \frac{x}{2^{n-1}} \right) - f\left( \frac{x}{2^n} \right) = \frac{x}{2^n} \]
3. Sum the series: \[ f(x) - f\left( \frac{x}{2^n} \right) = x \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^n} \right) \] \[ f(x) - f\left( \frac{x}{2^n} \right) = x \left( 1 - \frac{1}{2^n} \right) \]
4. Take the limit as $n \to \infty$: \[ G(x) = \lim_{n \to \infty} \left( f(x) - f\left( \frac{x}{2^n} \right) \right) = x \] 5. Calculate $\sum_{r=1}^{10} G(r^2)$: \[ \sum_{r=1}^{10} G(r^2) = \sum_{r=1}^{10} r^2 = 1^2 + 2^2 + 3^2 + \cdots + 10^2 \] \[ \sum_{r=1}^{10} r^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \] Therefore, the correct answer is (2) 385.