De la même manière on trouvera que l’on a,
étant encore égal à
![{\displaystyle {\begin{aligned}&\Pi \ \,(a_{2},a_{1})={\frac {q(\mathrm {B} )-2(\mathrm {A} )}{2a_{2}^{3}}},\\&\Pi _{1}(a_{2},a_{1})={\frac {q(\mathrm {C} )-2(\mathrm {B} )+2q(\mathrm {A} )}{2a_{2}^{3}}},\\&\Pi _{2}(a_{2},a_{1})={\frac {q(\mathrm {D} )-2(\mathrm {C} )+q(\mathrm {B} )}{2a_{2}^{3}}},\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b3ed3f459bb8299ea7de78ea0843e6a1a93af6c)
d’où, en faisant
![{\displaystyle {\begin{aligned}&\mathrm {Q} \ \,={\frac {q(\mathrm {B} )-2(\mathrm {A} )}{2}},\\&\mathrm {Q} _{1}={\frac {q(\mathrm {C} )-2(\mathrm {B} )+2(\mathrm {A} )}{2}},\\&\mathrm {Q} _{2}={\frac {q(\mathrm {D} )-2(\mathrm {C} )+q(\mathrm {B} )}{2}},\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a7c35a1184af022a0de61549d8475d4a2407cd9)
on tire
![{\displaystyle {\begin{aligned}&{\breve {\Pi }}\ \,(a_{2},a_{1})=3{\frac {q\mathrm {Q} _{1}-2\mathrm {Q} }{2}}-\mathrm {A} ,\\&{\breve {\Pi }}_{1}(a_{2},a_{1})=3{\frac {q\mathrm {Q} _{2}-2\mathrm {Q} _{1}+2q\mathrm {Q} }{2}}-\mathrm {B} ,\\&{\breve {\Pi }}_{2}(a_{2},a_{1})=3{\frac {q\mathrm {Q} _{3}-2\mathrm {Q} _{2}+q\mathrm {Q} _{1}}{2}}-\mathrm {C} ,\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9b6a82c6b77526216a475944b1b5856cafa3e83)
expressions qui serviront aussi pour les quantités
en faisant successivement ![{\displaystyle q={\frac {a_{1}}{a_{3}}},\ q={\frac {a_{2}}{a_{3}}},\ldots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4cf7381db46bec50acb1aa5bb3604cf46ca5da6c)
LXXXI.
Ayant donc fait le calcul de ces différentes quantités, j’ai trouvé les valeurs suivantes :