[tex]\it \dfrac{1}{k}-\dfrac{1}{k+2} =\dfrac{2}{k(k+2)} \Big|_{\cdot\dfrac{1}{2}} \Rightarrow \dfrac{1}{2}\cdot(\dfrac{1}{k}-\dfrac{1}{k+2} ) =\dfrac{1}{k(k+1)} \Rightarrow
\\\;\\ \\\;\\
\Rightarrow \dfrac{1}{k(k+1)} = \dfrac{1}{2}\cdot \left(\dfrac{1}{k}-\dfrac{1}{k+2} \right) \ \ \ \ \ (*)[/tex]
[tex]\it \dfrac{1}{1\cdot3} = \dfrac{1}{2}\cdot \left(\dfrac{1}{1}-\dfrac{1}{3} \right)
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\dfrac{1}{3\cdot5} = \dfrac{1}{2}\cdot \left(\dfrac{1}{3}-\dfrac{1}{5} \right)
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\vdots\\
\dfrac{1}{2013\cdot2015} = \dfrac{1}{2}\cdot \left(\dfrac{1}{2013}-\dfrac{1}{2015} \right) [/tex]
Numărul din enunț devine:
[tex]\it a=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3} -\dfrac{1}{5}+\ ...\ +\dfrac{1}{2013}-\dfrac{1}{2015}\right) =\dfrac{1}{2} \left(1-\dfrac{1}{2015}\right)=
\\\;\\ \\\;\\
=\dfrac{1}{2}\cdot\dfrac{2014}{2015} =\dfrac{1007}{2015}[/tex]
[tex]\it 0 \ \textless \ \dfrac{1007}{2015} \ \textless \ 1 \Rightarrow 0\ \textless \ a\ \textless \ 1 \Rightarrow [a] =0[/tex]
