Question:

$\textcolor{green}{\textbf{Key Idea}}$ Rewrite the numerator as \[ \sum_{k=1}^{10}\left(1-\cos(kx)\right). \] Then use \[ 1-\cos(kx)\sim \frac{k^2x^2}{2} \qquad (x\to 0). \] So the limit becomes \[ \frac{1}{2}\sum_{k=1}^{10}k^2. \] $\textcolor{blue}{\textbf{Solution}}$ We need \[ \lim_{x\to 0} \frac{10-\sum_{k=1}^{10}\cos(kx)}{x^2}. \] Write \[ 10-\sum_{k=1}^{10}\cos(kx) =\sum_{k=1}^{10}\left(1-\cos(kx)\right). \] Using \[ 1-\cos u\sim \frac{u^2}{2} \qquad (u\to 0), \] we get \[ 1-\cos(kx)\sim \frac{k^2x^2}{2}. \] Therefore \[ \lim_{x\to 0} \frac{10-\sum_{k=1}^{10}\cos(kx)}{x^2} =\frac{1}{2}\sum_{k=1}^{10}k^2. \] Now \[ \sum_{k=1}^{10}k^2 =\frac{10\cdot 11\cdot 21}{6} =385. \] Hence \[ \frac{a}{b}=\frac{385}{2}. \] Since \(385\) and \(2\) are co-prime, \[ a+b=385+2=387. \] \[ \boxed{387} \] $\textcolor{red}{\textbf{Key Trap}}$ The constant term cancels exactly: \[ 10-\sum_{k=1}^{10}1=0. \] So the first non-zero contribution is the \(x^2\)-term from the cosine expansion. If this cancellation is missed, the limit appears incorrectly divergent.