Questions mathématiques diverses

Question 26:
Sans utiliser un raisonnement par récurrence, montrer que pour tout entier naturel n;
$\dfrac{1}{5!}+\dfrac{1!}{6!}+\dfrac{2!}{7!}+...(n\;termes)=\dfrac{1}{4}\left({\dfrac{1}{4!}-\dfrac{n!}{(n+4)!}}\right)$

Réponse 26:
$S_n=\sum\limits_{k=1}^{n}{\dfrac{(k-1)!}{(k+4)!}}$
On pose $x_k=\dfrac{(k-1)!}{(k+4)!}$ $\Longrightarrow x_{k+1}=\dfrac{k!}{(k+5)!}$
$kx_k-(k+5)x_{k+1}$ $=\dfrac{k!}{(k+4)!}-\dfrac{k!}{(k+4)!}=0$ $\Longrightarrow kx_k=(k+5)x_{k+1}$
$kx_k-(k+1)x_{k+1}$ $=(k+5)x_{k+1}-(k+1)x_{k+1}$ $=4x_{k+1}$ $\Longrightarrow x_{k+1}=\dfrac{1}{4}\left({kx_k-(k+1)x_{k+1}}\right)$
$$\begin{align*} S_n&=\sum\limits_{k=1}^{n}{x_k}\\ &=x_1+\sum\limits_{k=2}^{n}{x_k}\\ &=\dfrac{1}{5!}+\sum\limits_{k=1}^{n-1}{x_{k+1}}\\ &=\dfrac{1}{5!}+\dfrac{1}{4}\sum\limits_{k=1}^{n-1}{\left({kx_k-(k+1)x_{k+1}}\right)}\\ &=\small{\dfrac{1}{5!}+\dfrac{1}{4}\left({(x_1-2x_2)+(2x_2-3x_3)+...+((n-1)x_{n-1}-nx_n)}\right)}\\ &=\dfrac{1}{5!}+\dfrac{1}{4}(x_1-nx_n)\\ &=\dfrac{1}{5!}+\dfrac{1}{4}\left({\dfrac{1}{5!}-\dfrac{n!}{(n+4)!}}\right)\\ &=\dfrac{1}{5!}\left({1+\dfrac{1}{4}}\right)-\dfrac{n!}{4(n+4)!}\\ &=\dfrac{1}{5!}\times\dfrac{5}{4}-\dfrac{n!}{4(n+4)!}\\ &=\dfrac{1}{4}\left({\dfrac{1}{4!}-\dfrac{n!}{(n+4)!}}\right) \end{align*}$$