Vom analiza una dintre fracții. O alegem pe a doua:
[tex]\it \dfrac{2}{5\cdot9} =\dfrac{a}{5}-\dfrac{b}{9}\ \ \ \ \ (*) \\ \\ \\ \dfrac{^{9)}1}{\ 5}-\dfrac{^{5)}1}{\ 9}=\dfrac{9-5}{5\cdot9}=\dfrac{4}{5\cdot9}=\dfrac{2^2}{5\cdot9}\\ \\ \\ Deci,\ \ \dfrac{2}{5\cdot9}= \dfrac{2}{2}\cdot \dfrac{2}{5\cdot9}= \dfrac{1}{2}\cdot \dfrac{2^2}{5\cdot9}= \dfrac{1}{2}\cdot\Big( \dfrac{1}{5}- \dfrac{1}{9}\Big)[/tex]
Să alegem acum ultima fracție din membrul stâng.
[tex]\it \dfrac{1}{2^n+1}-\dfrac{1}{2^{n+1}+1}=\dfrac{2^{n+1}+1-2^n-1}{(2^n+1)(2^{n+1}+1)}=\dfrac{2^{n+1}-2^n}{(2^n+1)(2^{n+1}+1)} =\\ \\ \\ =\dfrac{2^n(2-1)}{(2^n+1)(2^{n+1}+1)}=\dfrac{2^n}{(2^n+1)(2^{n+1}+1)}=\dfrac{2\cdot2^{n-1}}{(2^n+1)(2^{n+1}+1)}[/tex]
Așadar, membrul stâng se poate scrie:
[tex]\it \dfrac{1}{2}\Big(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\ ...\ +\dfrac{1}{2^n+1}-\dfrac{1}{2^{n+1}+1}\Big)=\dfrac{1}{2}\Big(\dfrac{1}{3}-\dfrac{1}{2^{n+1}+1}\Big)=\\ \\ \\ =\dfrac{1}{2}\cdot\dfrac{2^{n+1}+1-3}{3(2^{n+1}+1)}=\dfrac{1}{2}\cdot\dfrac{2^{n+1}-2}{3(2^{n+1}+1)}=\dfrac{1}{2}\cdot\dfrac{2(2^{n}-1)}{3(2^{n+1}+1)}=\dfrac{2^{n}-1}{3(2^{n+1}+1)}[/tex]
Acum, egalitatea din enunț devine:
[tex]\it \dfrac{2^{n}-1}{3(2^{n+1}+1)}=\dfrac{2^{2014}-1}{3(2^{2015}+1)} \Rightarrow 2^n-1=2^{2014}-1 \Rightarrow 2^n=2^{2014}\Rightarrow\\ \\ \\ \Rightarrow n=2014[/tex]
