Question

VA ROG AJUTOR
S = 1 supra 3 + 1 supra 3^2 +….+ 1 supra 3^100. Aratati ca S< 1 supra 2

Answer 1

Răspuns:

Explicație pas cu pas:

$S=\frac{1}{3} +(\frac{1}{3})^2+(\frac{1}{3} )^3+...+(\frac{1}{3} )^1^0^0 \ \ |\cdot \frac{1}{3} \\\\\frac{1}{3} \cdot S=(\frac{1}{3})^2+(\frac{1}{3} )^3+...+(\frac{1}{3} )^1^0^0+(\frac{1}{3})^1^0^1\\\\\frac{1}{3}\cdot S=S-\frac{1}{3}+(\frac{1}{3})^1^0^1\\\\\frac{2}{3} \cdot S=\frac{1}{3}-(\frac{1}{3})^1^0^1\ \ \ |\cdot\frac{3}{2} \\\\S=\frac{1}{2} -(\frac{1}{3})^1^0^0\cdot \frac{1}{2} <\frac{1}{2}$