trouvés
![{\displaystyle {\begin{aligned}\sideset {_{r}}{_{n}}{\mathrm {A} }=&\pm {\frac {n^{n-1}}{\nabla (r-1)\nabla (n-r)}},\\\sideset {_{r}}{_{n}}{\mathrm {B} }=&\mp {\frac {n^{n-2}a}{\nabla (r-2)\nabla (n-r)}},\\\sideset {_{r}}{_{n}}{\mathrm {C} }=&\mp {\frac {n^{n-3}a^{2}}{1.2}}\left[{\frac {1}{\nabla (r-2)\nabla (n-r-1)}}-{\frac {1}{\nabla (r-3)\nabla (n-r)}}\right],\\\sideset {_{r}}{_{n}}{\mathrm {D} }=&\mp {\frac {n^{n-4}a^{3}}{1.2.3}}\left[{\frac {1}{\nabla (r-2)\nabla (n-r-2)}}\right.\\&\qquad \qquad \quad \left.-{\frac {4}{\nabla (r-3)\nabla (n-r-1)}}-{\frac {1}{\nabla (r-4)\nabla (n-r)}}\right],\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c467ee35c68ba54c86bc75cef6d6353e107bb5d7)
le signe supérieur ayant lieu si
est impair, et l’inférieur s’il est pair. De là, je conclus que l’on a généralement
![{\displaystyle {\begin{aligned}\sideset {_{r}}{_{n}}{\mathrm {\overset {q}{T}} }&=\mp {\frac {n^{n-q}a^{q-1}}{\nabla (q-1)}}\\&\times \left[{\frac {1}{\nabla (r-2)\nabla (n-r-q+2)}}+{\frac {\mathrm {M} _{q}}{\nabla (r-3)\nabla (n-r-q+3)}}\right.\\&+\left.{\frac {\sideset {^{1}}{_{q}}{\mathrm {M} }}{\nabla (r-4)\nabla (n-r-q+4)}}+\ldots +{\frac {\sideset {^{q-3}}{_{q}}{\mathrm {M} }}{\nabla (r-q)\nabla (n-r)}}\right],\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d814d7d6db7d67c1b2288a2e87e79e6560764f86)
expression dans laquelle il faut déterminer
Pour y parvenir, j’observe que cette valeur de
ne peut commencer à exister que lorsque
or on a
![{\displaystyle \sideset {_{3}}{_{q}}{\mathrm {\overset {q}{T}} }=-{\frac {a^{q-1}\mathrm {M} _{q}}{\nabla (q-1)}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee3294892d88d7b3dbe6595cffc46fe2e2322f0e)
d’ailleurs, l’équation
donne
![{\displaystyle {\begin{aligned}\sideset {_{3}}{_{q}}{\mathrm {\overset {q}{T}} }=&\sideset {_{2}}{_{q}}{\mathrm {A} }\left({\frac {a}{q}}\right)^{q-1}+\sideset {_{2}}{_{q}}{\mathrm {B} }\left({\frac {a}{q}}\right)^{q-2}+\ldots +\sideset {_{2}}{_{q}}{\mathrm {\overset {q}{T}} }\\=&a^{q-1}\left[-{\frac {1}{\nabla (q-1)}}+{\frac {1}{\nabla (q-2)}}+{\frac {1}{1.2.\nabla (q-3)}}+\ldots +{\frac {1}{\nabla (q-1)}}\right]\\=&{\frac {a^{q-1}}{\nabla (q-1)}}\left[1+{\frac {q-1}{1}}+{\frac {(q-1)(q-2)}{1.2}}+\ldots +1-q\right]\\=&{\frac {a^{q-1}}{\nabla (q-1)}}\left(2^{q-1}-q\right)\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/476f2baf4d16157ab8c8934ac162d371c9c6013d)