Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 8.djvu/181

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donc, puisque $\sideset {_{2,2}}{_{4}}y=1,$

\sideset {_{2}}{_{2}}{\mathrm {L} }={\frac {1}{r^{2}}}-{\frac {2p}{r^{2}}}(1-q)-{\frac {2q}{r^{2}}}(1-p)-{\frac {2pq}{r^{2}}}\,;

de plus, $\mathrm {C}$ égale $\sideset {_{2}}{_{2}}{\mathrm {L} }$ dans l’expression de $\sideset {_{2}}{_{n}}{\mathrm {L} }.$

On trouvera pareillement

{\begin{aligned}\sideset {_{3}}{_{n}}{\mathrm {L} }&={\frac {q^{n-3}}{r^{n-3}}}{\frac {p^{2}}{q^{2}}}\left({\frac {1-q}{r}}\right)^{2}{\frac {(n-2)(n-1)}{1.2}}\\&+{\frac {q^{n-2}}{r^{n-2}}}{\frac {p}{q}}\sideset {_{2}}{_{2}}{\mathrm {L} }(n-1)\\&+\mathrm {C} {\frac {r^{n-1}}{q^{n-1}}},\end{aligned}}

$\mathrm {C}$ étant une constante arbitraire ; or, posant $n=1,\ \sideset {_{3}}{_{n}}{\mathrm {L} }=\left({\frac {1-p}{r}}\right)^{2}\,;$ donc

\mathrm {C} =\left({\frac {1-p}{r}}\right)^{2}\,;

partant,

{\begin{aligned}\sideset {_{3}}{_{n}}{\mathrm {L} }&={\frac {q^{n-3}}{r^{n-3}}}{\frac {p^{2}}{q^{2}}}\left({\frac {1-q}{r}}\right)^{2}{\frac {(n-2)(n-1)}{1.2}}\\&+{\frac {q^{n-2}}{r^{n-2}}}{\frac {p}{q}}\sideset {_{2}}{_{2}}{\mathrm {L} }(n-1)\\&+\left({\frac {1-p}{r}}\right)^{2}{\frac {q^{n-1}}{p^{n-1}}},\end{aligned}}

et généralement on aura

{\begin{aligned}\sideset {_{m}}{_{n}}{\mathrm {L} }&={\frac {q^{n-3}p^{m-1}}{r^{m+n-4}}}\left({\frac {1-q}{r}}\right)^{2}{\frac {(n-2)(n-1)\ldots (n+m-4)}{1.2.3\ldots (m-1)}}\\&+{\frac {q^{n-2}p^{m-2}}{r^{m+n-4}}}\sideset {_{2}}{_{2}}{\mathrm {L} }{\frac {(n-1)\ldots (n+m-4)}{1.2.3\ldots (m-2)}}\\&+{\frac {q^{n-1}p^{m-3}}{r^{m+n-4}}}\left({\frac {1-p}{r}}\right)^{2}{\frac {n\ldots (n+m-4)}{1.2.3\ldots (m-3)}}.\end{aligned}}

On a ensuite

\sideset {_{2}}{_{n}}{\mathrm {K} }={\frac {q^{n-4}}{r^{n-4}}}{\frac {p}{r}}\left({\frac {1-q}{r}}\right)^{3}+{\frac {q}{r}}\sideset {_{2}}{_{n-1}}{\mathrm {K} }\,;

partant,

\sideset {_{2}}{_{n}}{\mathrm {K} }={\frac {q^{n-4}}{r^{n-4}}}{\frac {p}{r}}\left({\frac {1-q}{r}}\right)^{3}(n-3)+\mathrm {C} {\frac {q^{n-3}}{r^{n-8}}}\,;