Questions mathématiques diverses

Question 76:
Soit $n$ un entier naturel.
Déterminer l'expression $X_n$ telle que $\dfrac{1^4}{1.3}+\dfrac{2^4}{3.5}+\dfrac{3^4}{5.7}+...+\dfrac{n^4}{(2n-1)(2n+1)}=\dfrac{1}{48}X_n+\dfrac{n}{16(2n+1)}$

Réponse 76:

$\dfrac{n^4}{(2n-1)(2n+1)}=\dfrac{4n^2+1}{16}+\dfrac{1}{32}\left({\dfrac{1}{2n-1}-\dfrac{1}{2n+1}}\right)$
$\sum\limits_{k=1}^{n}{\left({\dfrac{1}{2k-1}-\dfrac{1}{2k+1}}\right)}$ $=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}$ $=1-\dfrac{1}{2n+1}=\dfrac{2n}{2n+1}$

$$\begin{align*} S_n&=\sum\limits_{k=1}^{n}{\dfrac{k^4}{(2k-1)(2k+1)}}\\ &=\sum\limits_{k=1}^{n}{\dfrac{4k^2+1}{16}+\dfrac{1}{32}\left({\dfrac{1}{2k-1}-\dfrac{1}{2k+1}}\right)}\\ &=\dfrac{1}{4}\sum\limits_{k=1}^{n}{k^2}+\sum\limits_{k=1}^{n}{\dfrac{1}{16}}+\dfrac{1}{32}\dfrac{2n}{2n+1}\\ &=\dfrac{1}{4}\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n}{16}+\dfrac{n}{16(2n+1)}\\ &=\dfrac{2n^3+3n^2+n}{24}+\dfrac{3n}{48}+\dfrac{n}{16(2n+1)}\\ &=\dfrac{4n^3+6n^2+5n}{48}+\dfrac{n}{16(2n+1)} \end{align*}$$ Donc $X_n=4n^3+6n^2+5n$