[size=85][math]1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{k^2}+...=\frac{\pi^2}{6}[/math][br]A: [i]n[br][/i]B: [math]\frac{1}{n^2}[/math][br]C:[math]\sum_{k=1}^n\frac{1}{k^2}[/math][br]D: [math]\sqrt{6C}[/math][/size]

[size=85][math]1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...+\left(-1\right)^{n+1}\cdot\frac{1}{2n-1}+...=\frac{\pi}{4}[/math][br]A: [i]n[br][/i]B: 2[i]n[/i]-1 [br]C: [math]\frac{1}{2n-1}[/math][/size]

[size=85]D: [math]1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...+\left(-1\right)^{k+1}\cdot\frac{1}{2k-1}[/math][br]E: 4D[/size]

[size=85][url=https://hu.wikipedia.org/wiki/P%C3%AD_(sz%C3%A1m)#%7F'%22%60UNIQ--postMath-0000003D-QINU%60%22'%7F-t_tartalmaz%C3%B3_k%C3%A9pletek]Viete-féle végtelen szorzat[br][math]\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot...=\frac{2}{\pi}[/math][/url][/size]