Suma: 1+[tex] \frac{1}{2}+ \frac{1}{ 2^{2} } + \frac{1}{ 2^{3} }+...+ \frac{1}{ 2^{n+1} }= \frac{4095}{2048} [/tex]. Termeni sunt in progresie geometrica , cu ratia q=[tex] \frac{1}{2} [/tex] deci:
[tex] \frac{1- ( \frac{1}{2}) ^{n+2} }{1- \frac{1}{2} } = \frac{4095}{2048} [/tex]. Aducem la o forma mai simpla , egalam produsul mezilor cu la extremilor, si notam [tex] 2^{n}=y [/tex], obtinem ecuatia 2y=2048, adica y=1024, deci [tex] 2^{n}= 2^{10} [/tex], de unde n=10.
