Bonsoir,
Ex 3:
[tex]u_1=\dfrac{5}{2}\\u_{n+1}=\dfrac{5u_n-4}{2u_n-1}\\On\ pose\ v_n=\dfrac{u_n-2}{u_n-1}\\v_{n+1}=\dfrac{u_{n+1}-2}{u_{n+1}-1}\\\\=\dfrac{\dfrac{5u_n-4}{2u_n-1}-2}{\dfrac{5u_n-4}{2u_n-1}-1}\\\\=\dfrac{5u_n-4-4u_n+2}{5u_n-4-2u_n+1}\\\\=\dfrac{u_n-2}{3u_n-3}\\\\=\dfrac{1}{3} \dfrac{u_n-2}{u_n-1}\\\\\boxed{v_{n+1}=\dfrac{1}{3}v_n}\\\\v_1=\dfrac{u_1-2}{u_1-1}=\dfrac{1}{3}\\\\v_1=v_0*\dfrac{1}{3}\ \Rightarrow\ v_0=1\\\\v_n=1*(\dfrac{1}{3})^n\\\\[/tex]
[tex]\dfrac{u_n-2}{u_n-1}=\dfrac{1}{3^n}\\\\u_n-1=3^n(u_n-2)\\\\u_n(3^n-1)=-1+2*3^n\\\\u_n=1+\dfrac{3^n}{3^n-1}\\\\u_n=2+\dfrac{1}{3^n-1}\\\\[/tex]
ex 4:
[tex]u_0=4\\u_{n+1}=0.5*n-1\\-2\leq4\ \\\\Soit \ u_n \geq -2 \ vraie\ \\\\0.5*u_n\geq -1\\\\0.5 u_n -1\geq-2\\\\u_{n+1}\geq-2\\\\[/tex]
