27020 - Revision history

Andrei.Horvat at 10:50, 19 October 2023

2023-10-19T10:50:09Z

← Older revision		Revision as of 10:50, 19 October 2023
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	'''Soluție:'''		'''Soluție:'''

	Fie <math> a_n </math> coeficientul lui <math> X^n </math> din ~~rezolvarea~~ lui		Fie <math> a_n </math> coeficientul lui <math> X^n </math> din dezvoltarea lui

	<math display="block"> P(X) = \left( X + \frac{1}{2}\right)^{2n} = \left(X(1+X) + \frac{1}{4}\right)^n = \sum_{k=0}^n C_n^k X^{n-k} (~~! -~~ X)^{n-k} \left.\frac{1}{4^k}\right. .</math>		<math display="block"> P(X) = \left( X + \frac{1}{2}\right)^{2n} = \left(X(1+X) + \frac{1}{4}\right)^n = \sum_{k=0}^n C_n^k X^{n-k} (1+ X)^{n-k} \left.\frac{1}{4^k}\right. .</math>

	Avem <math> a_n = \left.\frac{1}{2^n}\right. C_{2n}^n </math>, iar pe de altă parte,		Avem <math> a_n = \left.\frac{1}{2^n}\right. C_{2n}^n </math>, iar pe de altă parte,
	<math display="block"> a_n = C_n^0 \cdot C_n^0 + C_n^1 \cdot C_{n-1}^1 \left.\frac{1}{4}\right.		<math display="block"> a_n = C_n^0 \cdot C_n^0 + C_n^1 \cdot C_{n-1}^1 \left.\frac{1}{4}\right.
	+ C_n^2 \cdot C_{n-2}^1\left.\frac{1}{4^2}\right. + ... = </math>		+ C_n^2 \cdot C_{n-2}^2\left.\frac{1}{4^2}\right. + ... = </math><math display="block"> = \sum_{k=0}^{\left[\frac{n}{2}\right]} C_n^k C_{n-k}^k \cdot \left.\frac{1}{4^k}\right. = n! \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{(k!)^2 (n-k)! 4^k},</math>
	<math display="block"> = \sum_{k=0}^{\left[\frac{n}{2}\right]} C_n^k C_{n-k}^k \cdot \left.\frac{1}{4^k}\right. = n! \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{(k!)^2 (n-k)! 4^k},</math>

	deci suma este egală cu <math> \left.\frac{(2n!)}{2^n(n!)^3}\right. .</math>		deci suma este egală cu <math> \left.\frac{(2n!)}{2^n(n!)^3}\right. .</math>

Nagy Lenard at 18:06, 18 October 2023

2023-10-18T18:06:25Z

← Older revision		Revision as of 18:06, 18 October 2023
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	'''27020 (Gheorghe Szöllösy)'''		'''27020 (Gheorghe Szöllösy)'''

	''Să se calculeze suma'' <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1		''Să se calculeze suma'' <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1.
	</math>.''		</math>''

	'''Soluție:'''		'''Soluție:'''

Nagy Lenard at 18:04, 18 October 2023

2023-10-18T18:04:49Z

← Older revision		Revision as of 18:04, 18 October 2023
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	'''27020 (Gheorghe Szöllösy)'''		'''27020 (Gheorghe Szöllösy)'''

	Să se calculeze suma <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1.		''Să se calculeze suma'' <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1
	</math>		</math>.''

	'''Soluție:'''		'''Soluție:'''

Nagy Lenard: Anularea modificării 6981 făcute de Nagy Lenard (Discuție)

2023-10-18T18:04:16Z

Anularea modificării 6981 făcute de Nagy Lenard (Discuție)

← Older revision		Revision as of 18:04, 18 October 2023
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	'''27020 (Gheorghe Szöllösy)'''		'''27020 (Gheorghe Szöllösy)'''

	~~\textit{italic}~~ Să se calculeze suma <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1.		Să se calculeze suma <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1.
	</math>		</math>

Nagy Lenard at 18:04, 18 October 2023

2023-10-18T18:04:01Z

← Older revision		Revision as of 18:04, 18 October 2023
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	'''27020 (Gheorghe Szöllösy)'''		'''27020 (Gheorghe Szöllösy)'''

	Să se calculeze suma <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1.		\textit{italic} Să se calculeze suma <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1.
	</math>		</math>

Nagy Lenard at 17:59, 18 October 2023

2023-10-18T17:59:22Z

← Older revision		Revision as of 17:59, 18 October 2023
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	'''27020 (Gheorghe Szöllösy)'''		'''27020 (Gheorghe Szöllösy)'''

	Să se calculeze suma <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1		Să se calculeze suma <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1.
	</math>		</math>

Nagy Lenard at 17:57, 18 October 2023

2023-10-18T17:57:17Z

← Older revision		Revision as of 17:57, 18 October 2023
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	<math display="block"> P(X) = \left( X + \frac{1}{2}\right)^{2n} = \left(X(1+X) + \frac{1}{4}\right)^n = \sum_{k=0}^n C_n^k X^{n-k} (! - X)^{n-k} \left.\frac{1}{4^k}\right. .</math>		<math display="block"> P(X) = \left( X + \frac{1}{2}\right)^{2n} = \left(X(1+X) + \frac{1}{4}\right)^n = \sum_{k=0}^n C_n^k X^{n-k} (! - X)^{n-k} \left.\frac{1}{4^k}\right. .</math>

	Avem <math> a_n = \left(\frac{1}{2^n}\right~~) C_2n~~^n </math>, iar pe de altă parte,		Avem <math> a_n = \left.\frac{1}{2^n}\right. C_{2n}^n </math>, iar pe de altă parte,
	<math display="block"> a_n = C_n^0 \cdot C_n^0 + C_n^1 \cdot C_{n-1}^1 \left.\frac{1}{4}\right.		<math display="block"> a_n = C_n^0 \cdot C_n^0 + C_n^1 \cdot C_{n-1}^1 \left.\frac{1}{4}\right.
	+ C_n^2 \cdot C_{n-2}^1\left.\frac{1}{4^2}\right. + ... = </math>		+ C_n^2 \cdot C_{n-2}^1\left.\frac{1}{4^2}\right. + ... = </math>
	<math display="block"> = \sum_{k=0}^{\left[\frac{n}{2}\right]} C_n^k C_(n-k)^k \cdot \left.\frac{1}{4^k}\right. = n! \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{(k!)^2 (n-k)! 4^k},</math>		<math display="block"> = \sum_{k=0}^{\left[\frac{n}{2}\right]} C_n^k C_{n-k}^k \cdot \left.\frac{1}{4^k}\right. = n! \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{(k!)^2 (n-k)! 4^k},</math>

	deci suma este egală cu <math> \left.\frac{(2n!}{2^n(n!)^3}\right. .</math>		deci suma este egală cu <math> \left.\frac{(2n!)}{2^n(n!)^3}\right. .</math>

Nagy Lenard at 17:54, 18 October 2023

2023-10-18T17:54:43Z

← Older revision		Revision as of 17:54, 18 October 2023
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	'''27020 (Gheorghe Szöllösy)'''		'''27020 (Gheorghe Szöllösy)'''

	Să se calculeze suma <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 ~~\cdot~~ (n-2k)!}, \quad n \geq 1		Să se calculeze suma <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1
	</math>		</math>

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	Fie <math> a_n </math> coeficientul lui <math> X^n </math> din rezolvarea lui		Fie <math> a_n </math> coeficientul lui <math> X^n </math> din rezolvarea lui

	<math display="block"> P(X) = (\left. X + \frac{1}{2}\right)^{2n} = (X(1+X) + ~~\left.~~\frac{1}{4}\right.)^n = \sum_{k=0}^n C_n^k X^{n-k} (! - X)^{n-k} \left(\frac{1}{4^k}\right)</math>		<math display="block"> P(X) = \left( X + \frac{1}{2}\right)^{2n} = \left(X(1+X) + \frac{1}{4}\right)^n = \sum_{k=0}^n C_n^k X^{n-k} (! - X)^{n-k} \left.\frac{1}{4^k}\right. .</math>

	Avem <math> a_n = \left(\frac{1}{2^n}\right) C_2n^n </math>, iar pe de altă parte,		Avem <math> a_n = \left(\frac{1}{2^n}\right) C_2n^n </math>, iar pe de altă parte,

Nagy Lenard at 17:51, 18 October 2023

2023-10-18T17:51:11Z

← Older revision		Revision as of 17:51, 18 October 2023
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	Fie <math> a_n </math> coeficientul lui <math> X^n </math> din rezolvarea lui		Fie <math> a_n </math> coeficientul lui <math> X^n </math> din rezolvarea lui

	<math display="block"> P(X) = \left(X + \frac{1}{2}\right)^{2n} = (X(1+X) + \left.\frac{1}{4}\right.)^n = \sum_{k=0}^n C_n^k (! - X)^{(n-k)} \left(\frac{1}{4^k}\right)</math>		<math display="block"> P(X) = (\left. X + \frac{1}{2}\right)^{2n} = (X(1+X) + \left.\frac{1}{4}\right.)^n = \sum_{k=0}^n C_n^k X^{n-k} (! - X)^{n-k} \left(\frac{1}{4^k}\right)</math>

	Avem <math> a_n = \left(\frac{1}{2^n}\right) C_2n^n </math>, iar pe de altă parte,		Avem <math> a_n = \left(\frac{1}{2^n}\right) C_2n^n </math>, iar pe de altă parte,

Nagy Lenard at 17:48, 18 October 2023

2023-10-18T17:48:36Z

← Older revision		Revision as of 17:48, 18 October 2023
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	Fie <math> a_n </math> coeficientul lui <math> X^n </math> din rezolvarea lui		Fie <math> a_n </math> coeficientul lui <math> X^n </math> din rezolvarea lui

	<math display="block"> P(X) = \left(X + \frac{1}{2}\right)^{2n} = ~~\left~~(X(1+X) + \~~lfloor~~ 1/4 ~~\rfloor~~\right)^n = \sum_{k=0}^n C_n^k (! - X)^{(n-k)} \left(\frac{1}{4^k}\right)</math>		<math display="block"> P(X) = \left(X + \frac{1}{2}\right)^{2n} = (X(1+X) + \left.\frac{1}{4}\right.)^n = \sum_{k=0}^n C_n^k (! - X)^{(n-k)} \left(\frac{1}{4^k}\right)</math>

	Avem <math> a_n = \left(\frac{1}{2^n}\right) C_2n^n </math>, iar pe de altă parte,		Avem <math> a_n = \left(\frac{1}{2^n}\right) C_2n^n </math>, iar pe de altă parte,
	<math display="block"> a_n = C_n^0 \cdot C_n^0 + C_n^1 \cdot C_(n-1)^1 \left.\frac{1}{4}\right.		<math display="block"> a_n = C_n^0 \cdot C_n^0 + C_n^1 \cdot C_{n-1}^1 \left.\frac{1}{4}\right.
	+ C_n^2 \cdot C_(n-2)^1\left.\frac{1}{4^2}\right. + ... = </math>		+ C_n^2 \cdot C_{n-2}^1\left.\frac{1}{4^2}\right. + ... = </math>
	<math display="block"> = \sum_{k=0}^{\left[\frac{n}{2}\right]} C_n^k C_(n-k)^k \cdot \left.\frac{1}{4^k}\right. = n! \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{(k!)^2 (n-k)! 4^k},</math>		<math display="block"> = \sum_{k=0}^{\left[\frac{n}{2}\right]} C_n^k C_(n-k)^k \cdot \left.\frac{1}{4^k}\right. = n! \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{(k!)^2 (n-k)! 4^k},</math>

	deci suma este egală cu <math> \left.\frac{(2n!}{2^n(n!)^3}\right. .</math>		deci suma este egală cu <math> \left.\frac{(2n!}{2^n(n!)^3}\right. .</math>