Question-197050

Question Number 197050 by Skabetix last updated on 06/Sep/23

Commented by TheHoneyCat last updated on 07/Sep/23

Maintenant, 2b Y_k =(((Σ_(k=1) ^n x^k )−n)/(x−1)) =(((Σ_(k=1) ^n x^k )−(Σ_(k=1) ^n 1))/(x−1)) =Σ_(k=1) ^n ((x^k −1)/(x−1)) d′ou lim_(x→1) Y_k =Σ_(k=1) ^n k=((n(n+1))/2)

$$\mathrm{Maintenant},\:\mathrm{2b} \\ $$$${Y}_{{k}} =\frac{\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{x}^{{k}} \right)−{n}}{{x}−\mathrm{1}} \\ $$$$=\frac{\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{x}^{{k}} \right)−\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1}\right)}{{x}−\mathrm{1}} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{x}^{{k}} −\mathrm{1}}{{x}−\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{d}'\mathrm{ou} \\ $$$$\mathrm{lim}_{{x}\rightarrow\mathrm{1}} {Y}_{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$ \\ $$

Commented by TheHoneyCat last updated on 07/Sep/23

Commencons par 2a mq ∀k∈N^∗ lim_(x→1) (((x^k −1)/(x−1)))=k X_k =(1/(x−1))(x^k −1) =(1/(x−1))(x−1)Σ_(κ=0) ^(k−1) x^κ =Σ_(κ=0) ^(k−1) x^κ d′ou: lim_(x→∞) X_k =Σ_(κ=0) ^(k−1) x^κ =Σ_(κ=0) ^(k−1) 1=k

$$\mathrm{Commencons}\:\mathrm{par}\:\mathrm{2a} \\ $$$$\mathrm{mq}\:\forall{k}\in\mathbb{N}^{\ast} \:\mathrm{lim}_{{x}\rightarrow\mathrm{1}} \left(\frac{{x}^{{k}} −\mathrm{1}}{{x}−\mathrm{1}}\right)={k} \\ $$$${X}_{{k}} =\frac{\mathrm{1}}{{x}−\mathrm{1}}\left({x}^{{k}} −\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{{x}−\mathrm{1}}\left({x}−\mathrm{1}\right)\underset{\kappa=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}{x}^{\kappa} \\ $$$$=\underset{\kappa=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}{x}^{\kappa} \\ $$$$\mathrm{d}'\mathrm{ou}: \\ $$$$\mathrm{lim}_{{x}\rightarrow\infty} {X}_{{k}} =\underset{\kappa=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}{x}^{\kappa} =\underset{\kappa=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\mathrm{1}={k} \\ $$

Commented by Skabetix last updated on 07/Sep/23

$${Merci}\:{beaucoup},\:{je}\:{vais}\:{essayer}\:{de}\:{comprendre}\:{votre}\:{correction} \\ $$

Commented by TheHoneyCat last updated on 07/Sep/23

j'ai un peu la flemme de réfléchir sur la 2c... mais si jamais ça bloque toujours, répondez à ce commentaire, ça me fera une notif' et je me pencherai sérieusement dessus.

Answered by Skabetix last updated on 06/Sep/23

I need help for the 2^(nd) question please Thanks in advance

$$\mathrm{I}\:{need}\:{help}\:{for}\:{the}\:\mathrm{2}^{{nd}} \:{question}\:{please} \\ $$$${Thanks}\:{in}\:{advance} \\ $$

Answered by witcher3 last updated on 12/Sep/23

2c n(n+1)=Σ_(k=1) ^n 2k ⇔Σ_(k=1) ^n ((2k(x^(k−1) −1))/(x−1))==Σ_(k=1) ^(n−1) ((2(k+1)(x^k −1))/((x−1))) =Σ_(k=1) ^(n−1) 2k(k+1)=(2/3)Σ_1 ^(n−1) ((k+1)^3 −k^3 −1) =((2n(n^2 −1))/3)

$$\mathrm{2c} \\ $$$$\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{2k} \\ $$$$\Leftrightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{2k}\left(\mathrm{x}^{\mathrm{k}−\mathrm{1}} −\mathrm{1}\right)}{\mathrm{x}−\mathrm{1}}==\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\frac{\mathrm{2}\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{k}} −\mathrm{1}\right)}{\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)} \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{2k}\left(\mathrm{k}+\mathrm{1}\right)=\frac{\mathrm{2}}{\mathrm{3}}\underset{\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\left(\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{k}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$$=\frac{\mathrm{2n}\left(\mathrm{n}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{3}} \\ $$$$ \\ $$

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