Răspuns:
[tex]f(x)=(\frac{1}{2})^{x}\\f(0)+f(1)+f(2)+...+f(n)+f(n+1)=(\frac{1}{2})^{0}+(\frac{1}{2})^{1}+(\frac{1}{2})^{2}+...+(\frac{1}{2})^{n}+(\frac{1}{2})^{n+1}=1*\frac{1-(\frac{1}{2})^{n+2} }{1-\frac{1}{2} } =2*(1-(\frac{1}{2})^{n+2})=\frac{4095}{2048},~deci~1-(\frac{1}{2})^{n+2}=\frac{4095}{4096} ,~deci~ (\frac{1}{2})^{n+2}=1-\frac{4095}{4096}=\frac{1}{4096}=(\frac{1}{2})^{12} ,~deci~ n+2=12, ~deci~n=10.[/tex]
Explicație pas cu pas:
[tex]f(0)*f(1)*f(2)*...*f(n+1)=(\frac{1}{2})^{0}*(\frac{1}{2})^{1}*(\frac{1}{2})^{2}*...*(\frac{1}{2})^{n+1}=(\frac{1}{2} )^{\frac{(n+1)*(n+2)}{2} }=(\frac{1}{2})^{66} ,~deci~(n+1)(n+2)=2*66[/tex]
⇒(n+1)(n+2)=132=11·12, ⇒n+1=11, ⇒n=11-1=10.
