Initialisation
Pour n=0, d'une part nous avons que u_{d,0}=0 et d'autre part :
\begin{array}{rcl}
r_d^n\times\dfrac{2d}{2-d}+\dfrac{2d}{d-2} & = & r_d^0\times\dfrac{2d}{2-d}+\dfrac{2d}{d-2} \
& = & \dfrac{2d}{2-d}+\dfrac{2d}{d-2} \
& = & -\dfrac{2d}{-(2-d)}+\dfrac{2d}{d-2} \
& = & -\dfrac{2d}{-2+d}+\dfrac{2d}{d-2} \
& = & -\dfrac{2d}{d-2}+\dfrac{2d}{d-2} \
& = & 0.\
\end{array}
L'égalité est donc vraie au rang n=0.
Hérédité
Supposons que pour un certain entier n\geq0 :
u_{d,n} = r_d^n\times\dfrac{2d}{2-d}+\dfrac{2d}{d-2}.
Montrons alors que :
u_{d,n+1} = r_d^{n+1}\times\dfrac{2d}{2-d}+\dfrac{2d}{d-2}.
On a :
\begin{array}{rcl}
u_{d,n+1} & = & r_d u_{d,n}+1 \
& = & r_d\left( r_d^n\times\dfrac{2d}{2-d}+\dfrac{2d}{d-2} \right)+1 \
& = & r_d^{n+1}\times\dfrac{2d}{2-d}+r_d\times\dfrac{2d}{d-2}+1 \
& = & r_d^{n+1}\times\dfrac{2d}{2-d}+\dfrac{d+2}{2d}\times\dfrac{2d}{d-2}+1 \
& = & r_d^{n+1}\times\dfrac{2d}{2-d}+\dfrac{d+2}{d-2}+1 \
& = & r_d^{n+1}\times\dfrac{2d}{2-d}+\dfrac{d+2}{d-2}+\dfrac{d-2}{d-2} \
& = & r_d^{n+1}\times\dfrac{2d}{2-d}+\dfrac{d+2+d-2}{d-2} \
& = & r_d^{n+1}\times\dfrac{2d}{2-d}+\dfrac{2d}{d-2}_\square\
\end{array}
Conclusion
D'après le principe de récurrence, pour tout entier n : u_{d,n} = r_d^n\times\dfrac{2d}{2-d}+\dfrac{2d}{d-2}.
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