![{\displaystyle \operatorname {f} _{2}(b,b,b)-\operatorname {f} _{2}(a,a,a)=(b-a)\left\{\operatorname {f} _{3}(a,b,b,b)+\operatorname {f} _{3}(a,a,b,b)+\operatorname {f} _{3}(a,a,a,b)\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e791015b5161fb2010c3c7820b2e32e4ca2249e3)
En continuant de la même manière, on aura en général
![{\displaystyle \operatorname {f} _{n-1}(b,b,b,\ldots b)-\operatorname {f} _{n-1}(a,a,a,\ldots a)=(b-a)\left\{{\begin{aligned}&\operatorname {f} _{n}(a,b,b,\ldots b,b,b)\\+&\operatorname {f} _{n}(a,a,b,\ldots b,b,b)\\+&\operatorname {f} _{n}(a,a,a,\ldots b,b,b)\\+&\ldots \ldots \ldots \ldots \ldots \\+&\operatorname {f} _{n}(a,a,a,\ldots a,b,b)\\+&\operatorname {f} _{n}(a,a,a,\ldots a,a,b)\end{aligned}}\right\}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b037f959015a83c11d46168a82445a35688db85)
si, dans cette formule, on fait
d’où
et qu’on change ensuite
en
, elle deviendra
![{\displaystyle \operatorname {f} _{n-1}(x+k,x+k,x+k,\ldots x+k,x+k)-\operatorname {f} _{n-1}(x,x,x,\ldots x,x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4cadee8ae9c101571eccc11a1441c73254175a57)
![{\displaystyle =k\left\{{\begin{aligned}&\operatorname {f} _{n}(x,x+k,x+k,\ldots x+k,x+k,x+k)\\+&\operatorname {f} _{n}(x,x,x+k,\ldots x+k,x+k,x+k)\\+&\operatorname {f} _{n}(x,x,x,\ldots x+k,x+k,x+k)\\+&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \\+&\operatorname {f} _{n}(x,x,x,\ldots x,x+k,x+k)\\+&\operatorname {f} _{n}(x,x,x,\ldots x,x+k)\end{aligned}}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc3430d3ab5b7edbdb3664dbe62d7d7e2c9baeca)
formule qui va nous servir tout-à-l’heure.
11. En conservant toujours les mêmes notations, on a, par la formule fondamentale,