Question

am nevoie de ajutor, va rog!! ​

Answer 1

Răspuns:

$\frac{2}{1*3} = \frac{1}{1} - \frac{1}{3} \\\\\frac{2}{3*5} = \frac{1}{3} - \frac{1}{5} \\\\\frac{2}{5*7} = \frac{1}{5} - \frac{1}{7}\\...\\\frac{2}{47*49} = \frac{1}{47} - \frac{1}{49}\\\\\\\frac{2}{49*51} = \frac{1}{49} - \frac{1}{51}\\\\daca le adunam observam ca o serie de terneni se reduc, ramanand \\\frac{1}{1} - \frac{1}{51} = \frac{51-1}{51} = \frac{50}{51}$

asadar, $\frac{50}{51}$ ∈($\frac{1}{2} , 1$)

Answer 2

$\it \dfrac{^{k+2)}1}{\ \ k}-\dfrac{^{k)}1}{\ k+2}=\dfrac{k+2-k}{k(k+2)}=\dfrac{2}{k(k+2)}\\ \\ \\ Deci,\ fiecare\ frac\c{\it t}ie\ din\ suma\ dat\breve a\ se\ descompune:\\ \\ \dfrac{1}{1}- \dfrac {1}{3}+ \dfrac {1}{3}- \dfrac {1}{5}+\ ...\ + \dfrac {1}{49}- \dfrac {1}{51}=1- \dfrac {1}{51} = \dfrac {50}{51} <1\ \ \ \ \ (1)$

$\it \dfrac {1}{2}< \dfrac {50}{51} \Rightarrow1\cdot51<2\cdot50 \Rightarrow 51<100\ \ (A)\\ \\ Deci,\ \ \dfrac {1}{2}< \dfrac {50}{51} \ \ \ \ \ (2) \\ \\ \\ (1),\ (2) \Rightarrow \dfrac {1}{2}< \dfrac {50}{51}< 1 \Rightarrow \dfrac{50}{51}\in\Big(\dfrac {1}{2},\ \ 1 \Big)$