Bonsoir,
Pour montrer voici un fichier en excel.
[tex]U_{n+2}- \dfrac{5}{2} U_{n+1}+U_n=0\\\\
r^2-\dfrac{5}{2}r+1=0\\\\
a\ pour\ solution\ r=2\ et\ r= \frac{1}{2} \\\\
U_n=k_1*2^n+k_2* \dfrac{1}{2^n} \\\\
Comme\ U_0=2\ \\
k_1*2^0+k_2* \dfrac{1}{2^0}=2\ ou\ k_1+k_2=2\\\\
Comme\ U_1=1\ \\
k_1*2^1+k_2* \dfrac{1}{2^1}=1\ ou\ 2*k_1+ \dfrac{k_2}{2}=1\\\\
k_1=0\ et\ k_2=2\\\\
U_n=2* \dfrac{1}{2^n} \\\\
\boxed{U_n=\dfrac{2}{2^n} }\\\\
[/tex]
[tex]V_0= \dfrac{U_1}{U_0} = \dfrac{1}{2} \\
V_1= \dfrac{U_2}{U_1} = \dfrac{\dfrac{1}{2}}{1}=\dfrac{1}{2} \\
V_2= \dfrac{U_3}{U_2} = \dfrac{\dfrac{1}{4}}{\dfrac{1}{2}}=\dfrac{1}{2} \\
[/tex]
[tex] V_{n+1}=\dfrac{U_{n+1}}{U_n} \\
V_{n+2}=\dfrac{U_{n+2}}{U_{n+1}} =\dfrac{ \dfrac{5}{2} U_{n+1}-U_n}{U_{n+1}} \\
V_{n+2}=\dfrac{5}{2}-\dfrac{1}{V_{n+1}}\\\\
Si\ V_{n+1}= \dfrac{1}{2}\ alors\
V_{n+2}=\dfrac{5}{2}-\dfrac{1}{\dfrac{1}{2} }=\dfrac{1}{2}
[/tex]