Question

Stabiliti daca numarul a= $\frac{2}{1x3} + \frac{2}{3x5}+...+ \frac{2}{49x51}$ apartine intervalului ($\frac{1}{2}$, 1)

Answer 1

[tex] \frac{2}{1\cdot3} + \frac{2}{3\cdot5}+...+ \frac{2}{49\cdot51}=\\ = \frac{1}{1} - \frac{1}{3} +\frac{1}{3} - \frac{1}{5}+...+\frac{1}{49} - \frac{1}{51}=\\ =\frac{1}{1}- \frac{1}{51}=\\ = \frac{50}{51} \\ Deoarece \frac{50}{51} \approx 0,98=>a\in( \frac{1}{2} ;1)[/tex]Scrie răspunsul tău aici