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Question-153696




Question Number 153696 by SANOGO last updated on 09/Sep/21
Answered by puissant last updated on 09/Sep/21
x_n =Σ_(k=1) ^n (1/(n^2 +k))  Calcul des 5 premiers termes..  → x_1 =(1/(1^2 +1))=(1/2)  → x_2 =(1/(2^2 +1))+(1/(2^2 +2))=(1/5)+(1/6)=((11)/(30))..  → x_3 =(1/(3^2 +1))+(1/(3^2 +2))+(1/(3^2 +3))=(1/(10))+(1/(11))+(1/(12))=((181)/(660)).  → x_4 =(1/(4^2 +1))+(1/(4^2 +2))+(1/(4^2 +3))+(1/(4^2 +4))=...  → x_5 =(1/(5^2 +1))+(1/(5^2 +2))+(1/(5^2 +3))+(1/(5^2 +4))+(1/(5^2 +1))=...    Sens de variation..    x_(n+1) −x_n <0 , (tu peux montrer par  recurrence)..  D′ou (x_n ) est decroissante...
$${x}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{k}} \\ $$$${Calcul}\:{des}\:\mathrm{5}\:{premiers}\:{termes}.. \\ $$$$\rightarrow\:{x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\rightarrow\:{x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} +\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{11}}{\mathrm{30}}.. \\ $$$$\rightarrow\:{x}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} +\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} +\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{11}}+\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{181}}{\mathrm{660}}. \\ $$$$\rightarrow\:{x}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} +\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} +\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} +\mathrm{4}}=… \\ $$$$\rightarrow\:{x}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} +\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} +\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} +\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} +\mathrm{1}}=… \\ $$$$ \\ $$$${Sens}\:{de}\:{variation}.. \\ $$$$ \\ $$$${x}_{{n}+\mathrm{1}} −{x}_{{n}} <\mathrm{0}\:,\:\left({tu}\:{peux}\:{montrer}\:{par}\right. \\ $$$$\left.{recurrence}\right).. \\ $$$${D}'{ou}\:\left({x}_{{n}} \right)\:{est}\:{decroissante}… \\ $$
Commented by SANOGO last updated on 09/Sep/21
tu es vraiment un champion
$${tu}\:{es}\:{vraiment}\:{un}\:{champion} \\ $$

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