[tex]\displaystyle\bf\\\sqrt{1-\frac{1}{2} +\frac{1}{4} -\frac{1}{8} +...}[/tex]
Под корнем бесконечно убывающая геометрическая прогрессия в которой :
[tex]\displaystyle\bf\\b_{1} =1\\\\q=-\frac{1}{2} \ \ tak \ kak \:\\\\\\1\cdot \Big(-\frac{1}{2} \Big)=-\frac{1}{2} \\\\\\\Big(-\frac{1}{2} \Big)\cdot\Big(-\frac{1}{2} \Big)=\frac{1}{4} \\\\\\\frac{1}{4} \cdot\Big(-\frac{1}{2} \Big)=-\frac{1}{8} \ ...\\\\\\S=\frac{b_{1} }{1-q} =\frac{1}{1-(-\frac{1}{2} )} =\frac{1}{1+\frac{1}{2} } =\frac{1}{\frac{3}{2} }=\frac{2}{3} \\\\\\\sqrt{1-\frac{1}{2} +\frac{1}{4} -\frac{1}{8} +...} =\sqrt{\frac{2}{3} } =\frac{\sqrt{2} }{\sqrt{3} } =\frac{\sqrt{6} }{3}[/tex]
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[tex]\displaystyle\bf\\\sqrt{1-\frac{1}{2} +\frac{1}{4} -\frac{1}{8} +...}[/tex]
Под корнем бесконечно убывающая геометрическая прогрессия в которой :
[tex]\displaystyle\bf\\b_{1} =1\\\\q=-\frac{1}{2} \ \ tak \ kak \:\\\\\\1\cdot \Big(-\frac{1}{2} \Big)=-\frac{1}{2} \\\\\\\Big(-\frac{1}{2} \Big)\cdot\Big(-\frac{1}{2} \Big)=\frac{1}{4} \\\\\\\frac{1}{4} \cdot\Big(-\frac{1}{2} \Big)=-\frac{1}{8} \ ...\\\\\\S=\frac{b_{1} }{1-q} =\frac{1}{1-(-\frac{1}{2} )} =\frac{1}{1+\frac{1}{2} } =\frac{1}{\frac{3}{2} }=\frac{2}{3} \\\\\\\sqrt{1-\frac{1}{2} +\frac{1}{4} -\frac{1}{8} +...} =\sqrt{\frac{2}{3} } =\frac{\sqrt{2} }{\sqrt{3} } =\frac{\sqrt{6} }{3}[/tex]