Christmas Streak 61/88: 2018 CSAT (Korean SAT) #30 of Liberal Arts

If we define $F(x) = \tfrac12x(1-x)$ for $0 \le x \le 1$ , then we can write $g(x) \; = \; 2^{-\lfloor x \rfloor}F(\{x\}) + x \hspace{2cm} x \ge 0$ where $\lfloor x \rfloor$ and $\{x\}$ are the integer and fractional parts of $x$ respectively. Then we have (assuming that $k$ is a positive integer greater than $5$ ) $h(x) \; = \; \frac{\varepsilon_{\lfloor x \rfloor}}{2^{\lfloor x \rfloor}} F(\{x \}) + x \hspace{2cm} x \ge 0$ where $\varepsilon_j \; = \; \left\{ \begin{array}{lll} -1 & \hspace{1cm} & 5 \le j \le k-1 \\ 1 & & \text{otherwise} \end{array} \right.$ and so $\begin{aligned} a_n \; = \; \int_0^n h(x)\, dx & = \; \tfrac12n^2 + \sum_{j=0}^{n-1} \frac{\varepsilon_j}{2^j} \int_0^1 F(x)\,dx \; = \; \tfrac12n^2 + \tfrac{1}{12}\sum_{j=1}^{n-1}2^{-j} - \tfrac16\sum_{j=5}^{k-1} 2^{-j} \\ & = \; \tfrac12n^2 + \tfrac16(1 - 2^{-n}) - \tfrac{1}{96}(1 - 2^{5-k}) \end{aligned}$ and hence $\lim_{n \to \infty} (2a_n - n^2) \; = \; \tfrac13 - \tfrac{1}{48}(1 - 2^{5-k}) \; = \; \tfrac{15}{48} + \tfrac13 2^{1-k}$ Since this limit equals $\tfrac{241}{768}$ , we deduce that $k = \boxed{9}$ .