Question

aratati ca 2/3<1/10+1/11+1/12+...........+1/27<9/5​

Answer 1

 

$\displaystyle\bf\\Trebuie~sa~aratam~ca:\\\\\frac{2}{3}<\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{26}+\frac{1}{27}<\frac{9}{5}\\\\\text{\bf Calculam numarul termenilor sumei de fractii.}\\\text{\bf Avem numitorii de la 10 pana la 27.}\\\\n=27-10+1=17+1=18~termeni~(fractii)$

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$\displaystyle\bf\\\text{\bf Ne ocupam de prima inegalitate:}\\\\\frac{1}{10}>\frac{1}{27}\\\\\frac{1}{11}>\frac{1}{27}\\\\\frac{1}{12}>\frac{1}{27}\\\\............\\\\\frac{1}{26}>\frac{1}{27}\\\\\frac{1}{27}=\frac{1}{27}\\\\\implies\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{26}+\frac{1}{27}>\underbrace{\frac{1}{27}+\frac{1}{27}+\frac{1}{27}+...+\frac{1}{27}+\frac{1}{27}}_{18~termeni}=\\\\=\frac{18}{27}=\frac{2}{3}$

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$\displaystyle\bf\\\implies~\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{26}+\frac{1}{27}>\frac{2}{3}\\\\\text{\bf Ne ocupam de a doua inegalitate:}\\\\\frac{1}{27}<\frac{1}{10}\\\\\frac{1}{26}<\frac{1}{10}\\\\..............\\\\\frac{1}{12}<\frac{1}{10}\\\\\frac{1}{11}<\frac{1}{10}\\\\\frac{1}{10}=\frac{1}{10}$

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$\displaystyle\bf\\\implies\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{26}+\frac{1}{27}<\underbrace{\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}+\frac{1}{10}}_{18~termeni}=\\\\ =\frac{18}{10}=\frac{9}{5}\\\\\\\implies\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{26}+\frac{1}{27}<\frac{9}{5}\\\\========================\\\\Rezulta~ca:\\\boxed{\bf\frac{2}{3}<\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{26}+\frac{1}{27}<\frac{9}{5}}$