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

Commented by TheHoneyCat last updated on 07/Sep/23

$Maintenant, 2 b$ $Y_{k} = \frac{(\underset{k = 1}{\sum^{n}} x^{k}) - n}{x - 1}$ $= \frac{(\underset{k = 1}{\sum^{n}} x^{k}) - (\underset{k = 1}{\sum^{n}} 1)}{x - 1}$ $= \underset{k = 1}{\sum^{n}} \frac{x^{k} - 1}{x - 1}$ $d^{'} ou$ $\lim_{x \to 1} Y_{k} = \underset{k = 1}{\sum^{n}} k = \frac{n (n + 1)}{2}$
Commented by TheHoneyCat last updated on 07/Sep/23

$Commencons par 2 a$ $mq \forall k \in N^{*} \lim_{x \to 1} (\frac{x^{k} - 1}{x - 1}) = k$ $X_{k} = \frac{1}{x - 1} (x^{k} - 1)$ $= \frac{1}{x - 1} (x - 1) \underset{κ = 0}{\sum^{k - 1}} x^{κ}$ $= \underset{κ = 0}{\sum^{k - 1}} x^{κ}$ $d^{'} ou :$ $\lim_{x \to \infty} X_{k} = \underset{κ = 0}{\sum^{k - 1}} x^{κ} = \underset{κ = 0}{\sum^{k - 1}} 1 = k$
Commented by Skabetix last updated on 07/Sep/23

$M e r c i b e a u c o u p, j e v a i s e s s a y e r d e c o m p r e n d r e v o t r e c o r r e c t i o n$
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

	$I n e e d h e l p f o r t h e 2^{n d} q u e s t i o n p l e a s e$ $T h a n k s i n a d v a n c e$
	Answered by witcher3 last updated on 12/Sep/23

	$2 c$ $n (n + 1) = \underset{k = 1}{\sum^{n}} 2 k$ $\Leftrightarrow \underset{k = 1}{\sum^{n}} \frac{2 k (x^{k - 1} - 1)}{x - 1} == \underset{k = 1}{\sum^{n - 1}} \frac{2 (k + 1) (x^{k} - 1)}{(x - 1)}$ $= \underset{k = 1}{\sum^{n - 1}} 2 k (k + 1) = \frac{2}{3} \underset{1}{\sum^{n - 1}} ({(k + 1)}^{3} - k^{3} - 1)$ $= \frac{2 n (n^{2} - 1)}{3}$
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