Question

Să demonstrăm că:
1:2²+1:3²+1:4²+...+1:2015²<2014:2015

Answer 1

$\frac{1}{2^2}\ \textless \ \frac{1}{1\cdot 2}=\frac{1}{1}-\frac{1}{2};\\\\\frac{1}{3^2}\ \textless \ \frac{1}{2\cdot 3}=\frac{1}{2}-\frac{1}{3};\\\\\frac{1}{4^2}\ \textless \ \frac{1}{3\cdot 4}=\frac{1}{3}-\frac{1}{4};\\\ldots\\\frac{1}{2014^2}\ \textless \ \frac{1}{2013\cdot 2014}=\frac{1}{2013}-\frac{1}{2014}\\\\\frac{1}{2015^2}\ \textless \ \frac{1}{2014\cdot 2015}=\frac{1}{2014}-\frac{1}{2015}\\\\Adun\u{a}m\;toate\;aceste\;2014\;inegalit\u{a}\c{t}i\;membru\;cu\;membru:\\\\\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots+\frac{1}{2015^2}\ \textless \ 1-\frac{1}{2015}=\frac{2014}{2015}$

Ceea ce trebuia demonstrat...

Green eyes.