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\frac{3}{4 \pi} \sqrt{4 \cdot x^2 12}\\ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}\\ \it{f}(x) = \frac{1}{\sqrt{x} x^2}\\ e^{i \pi} + 1 = 0\;

$ a + b = c $

\begin{eqnarray*} & & \frac{3}{4 \pi} \sqrt{4 \cdot x^2 12}\\ & & \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}\\ & & {\it f}(x) = \frac{1}{\sqrt{x} x^2}\\ & & e^{i \pi} + 1 = 0\; \end{eqnarray*}