${\displaystyle {\begin{aligned}\sum _{k=1}^{n}\sum _{i=1}^{n}\ln \left(i^{k}\right)&=\sum _{k=1}^{n}\sum _{i=1}^{n}k\ln(i)\\&=\sum _{k=1}^{n}k\sum _{i=1}^{n}\ln(i)\\&={\frac {n(n+1)}{2}}\ln(n!).\end{aligned}}}$

${\displaystyle {\begin{aligned}a)\,\sum _{1\leqslant i\leqslant j\leqslant n}{\frac {i}{j}}&=\sum _{j=1}^{n}{\frac {1}{j}}\sum _{i=1}^{j}i\\&=\sum _{j=1}^{n}{\frac {j+1}{2}}\\&={\frac {1}{4}}\left((n+1)(n+2)-1\times 2\right)\\&={\frac {n(n+3)}{4}}.\end{aligned}}}$

b) Soient ${\displaystyle S=\sum _{1\leq i<j\leq n}ij=\sum _{1\leq j<i\leq n}ij}$ et ${\displaystyle T=\sum _{1\leq i=j\leq n}ij}$.

${\displaystyle T={\frac {n(n+1)(2n+1)}{6}}}$ et ${\displaystyle 2S+T=\left({\frac {n(n+1)}{2}}\right)^{2}}$ donc ${\displaystyle \sum _{1\leq i\leq j\leq n}ij=S+T={\frac {n(n+1)^{2}}{2}}{\frac {n(n+1)(n+2)(3n+1)}{24}}}$.

c) Soient ${\displaystyle S=\sum _{1\leq i<j\leq n}i+j=\sum _{1\leq j<i\leq n}i+j}$ et ${\displaystyle T=\sum _{1\leq i=j\leq n}i+j}$.

${\displaystyle T=n(n+1)}$ et ${\displaystyle 2S+T=n^{2}(n+1)}$ donc ${\displaystyle \sum _{1\leq i\leq j\leq n}i+j=S+T={\frac {n(n+1)^{2}}{2}}}$.

a) Soient ${\displaystyle S=\sum _{1\leq i<j\leq n}ij=\sum _{1\leq j<i\leq n}ij}$ et ${\displaystyle T=\sum _{1\leq i=j\leq n}ij}$.

${\displaystyle T={\frac {1-a^{2n+2}}{1-a^{2}}}}$ et ${\displaystyle 2S+T=\left({\frac {1-a^{n+1}}{1-a}}\right)^{2}}$ donc ${\displaystyle \sum _{1\leq i\leq j\leq n}ij=S+T={\frac {(1-a^{n+1})(1-a^{n+2})}{(1-a)^{2}(1+a)}}}$.

b)

${\displaystyle {\begin{aligned}\sum _{1\leq i,j\leq n}\max(i,j)&=\sum _{k=1}^{n}k\left(1+\sum _{1\leq i<k}1+\sum _{1\leq j<k}1\right)\\&=\sum _{k=1}^{n}k\left(2k-1\right)\\&=2\sum _{k=1}^{n}k^{2}-\sum _{k=1}^{n}k\\&={\frac {n(n+1)(2n+1)}{3}}-{\frac {n(n+1)}{2}}\\&={\frac {n(n+1)(4n-1)}{6}}.\end{aligned}}}$

c) On peut raisonner comme dans la question précédente ou utiliser son résultat, en remarquant que ${\displaystyle \max(i,j)+\min(i,j)=i+j}$ et ${\displaystyle \sum _{1\leq i,j\leq n}(i+j)=n^{2}(n+1)}$, ce qui donne :

${\displaystyle \sum _{1\leq i,j\leq n}\min(i,j)=n^{2}(n+1)-{\frac {n(n+1)(4n-1)}{6}}={\frac {n(n+1)(2n+1)}{6}}}$.

d) On peut raisonner comme dans les deux questions précédente ou utiliser leurs résultats, en remarquant que ${\displaystyle |i-j|=\max(i,j)-\min(i,j)}$, ce qui donne : :${\displaystyle \sum _{1\leq i,j\leq n}|i-j|={\frac {n(n+1)(4n-1)}{6}}-{\frac {n(n+1)(2n+1)}{6}}={\frac {n(n+1)(n-1)}{3}}}$.

${\displaystyle a)\,\sum _{i=0}^{n}\sum _{j=0}^{p}u_{i,j}=\sum _{j=0}^{p}\sum _{i=0}^{n}u_{i,j}}$

${\displaystyle b)\,\sum _{i=0}^{n}\sum _{j=0}^{i}u_{i,j}=\sum _{j=0}^{n}\sum _{i=j}^{n}u_{i,j}}$

${\displaystyle c)\,\sum _{i=0}^{n}\sum _{j=i}^{p}u_{i,j}=\sum _{j=0}^{p}\sum _{i=0}^{\min(j,n)}u_{i,j}}$

${\displaystyle d)\,\sum _{i=0}^{\infty }\sum _{j=0}^{i}u_{i,j}=\sum _{j=0}^{\infty }\sum _{i=j}^{\infty }u_{i,j}}$

${\displaystyle e)\,\sum _{i=0}^{n-1}\sum _{j=i+1}^{\infty }u_{i,j}=\sum _{j=1}^{\infty }\sum _{i=0}^{\min(j,n)-1}u_{i,j}}$.

a)

${\displaystyle {\begin{aligned}\sum _{1\leq i\leq j\leq k\leq n}{\frac {ik}{j}}&=\sum _{j=1}^{n}{\frac {1}{j}}\sum _{i=1}^{j}i\sum _{k=j}^{n}k\\&={\frac {1}{4}}\sum _{j=1}^{n}(j+1)\left(n(n+1)-(j-1)j\right)\\&={\frac {1}{4}}\sum _{j=1}^{n}\left(-j^{3}+(n^{2}+n+1)j+n(n+1)\right)\\&={\frac {1}{4}}\left(-{\frac {n^{2}(n+1)^{2}}{4}}+(n^{2}+n+1){\frac {n(n+1)}{2}}+n^{2}(n+1)\right)\\&={\frac {n(n+1)(n^{2}+5n+2)}{16}}.\end{aligned}}}$

b) D'après l'exercice 4-2 b),

${\displaystyle \sum _{1\leq i\leq j\leq k}ij={\frac {k(k+1)(k+2)(3k+1)}{24}}}$

donc

${\displaystyle {\begin{aligned}\sum _{1\leq i\leq j\leq k\leq n}{\frac {ij}{k}}&=\sum _{k=1}^{n}{\frac {(k+1)(k+2)(3k+1)}{24}}\\&={\frac {1}{24}}\sum _{k=1}^{n}(3k^{3}+10k^{2}+9k+2)\\&={\frac {1}{24}}\left(3{\frac {n^{2}(n+1)^{2}}{4}}+10{\frac {n(n+1)(2n+1)}{6}}+9{\frac {n(n+1)}{2}}+2n\right)\\&={\frac {n(9n^{3}+58n^{2}+123n+98)}{288}}.\end{aligned}}}$

c) D'après l'exercice 4-2 a),

${\displaystyle \sum _{1\leq i\leq j\leq k}{\frac {i}{j}}={\frac {k(k+3)}{4}}}$

donc

${\displaystyle {\begin{aligned}\sum _{1\leq i\leq j\leq k\leq l\leq n}{\frac {i}{jkl}}&=\sum _{1\leq k\leq l\leq n}{\frac {k+3}{4l}}\\&=\sum _{l=1}^{n}{\frac {1}{4l}}\sum _{k=1}^{l}(k+3)\\&={\frac {1}{8}}\sum _{l=1}^{n}{\frac {1}{l}}\left((l+3)(l+4)-3\times 4\right)\\&={\frac {1}{8}}\sum _{l=1}^{n}(l+7)\\&={\frac {1}{16}}\left((n+7)(n+8)-7\times 8\right)\\&={\frac {n(n+15)}{16}}.\end{aligned}}}$

a) Nous allons décomposer cette somme en somme de sommes de termes constants.

${\displaystyle {\begin{aligned}\sum _{k=1}^{n^{2}}E\left({\sqrt {k}}\right)&=n+\sum _{i=1}^{n-1}\sum _{k=i^{2}}^{(i+1)^{2}-1}i\\&=n+\sum _{i=1}^{n-1}i(2i+1)\\&=2\sum _{i=1}^{n-1}i^{2}+\sum _{i=1}^{n-1}i+n\\&=2{\frac {(n-1)n(2n-1)}{6}}+{\frac {(n-1)n}{2}}+n\\&={\frac {n(4n^{2}-3n+5)}{6}}.\end{aligned}}}$

b) Même méthode que pour la question a) :

${\displaystyle {\begin{aligned}\sum _{k=1}^{n^{3}}E\left({\sqrt[{3}]{k}}\right)&=n+\sum _{i=1}^{n-1}\sum _{k=i^{3}}^{(i+1)^{3}-1}i\\&=n+\sum _{i=1}^{n-1}i\left((i+1)^{3}-i^{3}\right)\\&=n+\sum _{i=1}^{n-1}i\left(3i^{2}+3i+1\right)\\&=n+3\sum _{i=1}^{n-1}i^{3}+3\sum _{i=1}^{n-1}i^{2}+\sum _{i=1}^{n-1}i\\&=n+3{\frac {n^{2}(n-1)^{2}}{4}}+3{\frac {n(n-1)(2n-1)}{6}}+{\frac {n(n-1)}{2}}\\&={\frac {n(n+1)(3n^{2}-5n+4)}{4}}.\end{aligned}}}$