Réponse :
Bonsoir,
Explications étape par étape
[tex]u_0=1\\u_1=2\\u_{n+2}=1.5u_{n+1}-0.5u_n\\\\1a)\\v_n=u_{n+1}-u_n\\\\v_{n+1}=u_{n+2}-u_{n+1}\\=1.5u_{n+1}-0.5u_n-u_{n+1}\\=0.5u_{n+1}-0.5u_n\\=0.5(u_{n+1}-u_n)\\\\\boxed{v_{n+1}=\dfrac{1}{2} v_n}\\v_0=u_1-u_0=2-1=1\\\\1b)\\\boxed{v_n=\dfrac{1}{2^n} }\\[/tex]
2)
[tex]a)\\s_n=0.5*(1+\dfrac{1}{2} +\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{n-1}})\\\\=\dfrac{1}{2}*\dfrac{\dfrac{1}{2^n}-1}{\dfrac{1}{2}-1}\\\\s_n=1-\dfrac{1}{2^n}\\[/tex]
[tex]b)\\u_1-u_0=v_0=1\\\\u_2-u_1=v_1=\frac{1}{2} \\\\u_3-u_2=v_2=\dfrac{1}{2^2}\\\\u_4-u_3=v_2=\dfrac{1}{2^3}\\...\\u_{n+1}-u_n=v_n=\dfrac{1}{2^n}\\\\u_{n+1}-u_0=1+1-\dfrac{1}{2^n}\\\\\\u_{n+1}=3-\dfrac{1}{2^n}\\\\\boxed{u_{n}=3-\dfrac{1}{2^{n-1}}}\\[/tex]
