Question

Sa se determine termenii x2,x3,x4, și sa se exprime xn +2

Answer 1

Explicație pas cu pas:

$x_{2} = \dfrac{1}{2}x_{1} + 4 = - \dfrac{1}{2} + 4 = \dfrac{7}{2} \\ \implies x_{2} = \dfrac{8( {2}^{1} - 1) - 1}{ {2}^{1} }$

$x_{3} = \dfrac{1}{2}x_{2} + 4 = \dfrac{7}{4} + 4 = \dfrac{23}{4} \\ \implies x_{3} = \dfrac{8( {2}^{2} - 1) - 1}{ {2}^{2} }$

$x_{4} = \dfrac{1}{2}x_{3} + 4 = \dfrac{23}{8} + 4 = \dfrac{55}{8} \\ \implies x_{4} = \dfrac{8( {2}^{3} - 1) - 1}{ {2}^{3} }$

=>

$x_{n+2} = \dfrac{8( {2}^{n+1} - 1) - 1}{ {2}^{n+1} }$