Question

f(x)=$(\frac{1}{2}) ^{x }$

Sa se determine n∈N pentru care:

f(0) + f(1) + f(2)+.... f(n) + f(n+1) = $\frac{4095}{2048}$

Answer 1

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