This problem is a simple application of the sequential definition of continuity from ISI MStat 2015 PSB Problem 1 based on Telescopic Continuity.
Let \(f: R \rightarrow R\) be a function which is continuous at 0 and \(f(0)=1\)
Also assume that \(f\) satisfies the following relation for all \(x\) :
$$
f(x)-f(\frac{x}{2})=\frac{3 x^{2}}{4}+x
$$ Find \(f(3)\).
$$
f(3)-f(\frac{3}{2})=\frac{3 \times 3^{2}}{4}+3
$$
$$
f(\frac{3}{2})-f(\frac{\frac{3}{2}}{2})=\frac{3 \times \frac{3}{2}^{2}}{4}+\frac{3}{2}
$$
$$
f(\frac{3}{2^2})-f(\frac{\frac{3}{2^2}}{2})=\frac{3 \times \frac{3}{2^2}^{2}}{4}+\frac{3}{2^2}
$$
\( \cdots \)
$$
f(\frac{3}{2^n})-f(\frac{\frac{3}{2^n}}{2})=\frac{3 \times \frac{3}{2^n}^{2}}{4}+\frac{3}{2^n}
$$
Add them all up. That's the telescopic elegance.
$$
f(3)-f(\frac{3}{2^{n+1}})= \frac{3 \times 3^{2}}{4} \times \sum_{k = 0}^{n} \frac{1}{2^{2k}} + 3 \times \sum_{k = 0}^{n} \frac{1}{2^k} \rightarrow [*]
$$
Observe that \( a_n \to 0 \Rightarrow f(a_n) \to f(0) = 1\) since, \(f(x)\) is continuous at \(x=0\).
Hence take limit \( n \to \infty \) on \([*]\), and we get \( f(3) - f(0) = \frac{3 \times 3^{2}}{4} \times \frac{4}{3} + 3 \times 2 = 15 \).
