Andrei.Horvat at 17:56, 17 September 2025

2025-09-17T17:56:35Z

← Older revision		Revision as of 17:56, 17 September 2025
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	'''Soluție'''		'''Soluție'''

	Pentru orice număr natural <math>n</math>, cu <math> n\ge 2</math> are loc inegalitatea		Pentru orice număr natural <math>n</math>, cu <math> n\ge 2</math> are loc inegalitatea
	<math display="block"> \sqrt[n]{n!} = \sqrt[n]{1\cdot 2\cdot \ldots \cdot n} < \frac{1+2+\ldots+n}{n} = \frac{n+1}{2}.</math>		<math display="block"> \sqrt[n]{n!} = \sqrt[n]{1\cdot 2\cdot \ldots \cdot n} < \frac{1+2+\ldots+n}{n} = \frac{n+1}{2}.</math>Atunci<math display="block"> \sum_{i=2}^{n} \sqrt[i]{\left(i!\right)^2} < \sum_{i=2}^{n} \frac{\left(i+1\right)^2}{4}</math>Suma <math>S_n=\sum_{i=2}^{n} \frac{\left(i+1\right)^2}{4}</math> poate fi calculată prin mai multe metode. De exemplu, avem<math display="block">S_n = \frac{1}{4} \cdot \sum_{i=3}^{n+1} i^2 = \frac{1}{4} \left(\frac{n\left(n+1\right)\left(2n+1\right)}{6} - 5\right) = \frac{2n^3+9n^2+13n-24}{24}.</math>În concluzie, pentru orice <math>n\ge 2 </math> are loc inegalitatea<math display="block"> \sum_{i=2}^{n} \sqrt[i]{\left(i!\right)^2} < \frac{2n^3+9n^2+13n-24}{24} .</math>'''◎''' soluție oferită de '''D. Bărbosu'''
	Atunci
	<math display="block"> \sum_{i=2}^{n} \sqrt[i]{\left(I!\right)2} < \sum_{i=2}^{n} \frac{\left(i+1\right)^2}{4}</math>
	Suma <math>S_n=\sum_{i=2}^{n} \frac{\left(i+1\right)^2}{4}</math> poate fi calculată prin mai multe metode. De exemplu, avem
	<math display="block">S_n = \frac{1}{4} \cdot \sum_{i=3}^{n+1} i^2 = \frac{1}{4} \left(\frac{n\left(n+1\right)\left(2n+1\right)}{6} - 5\right) = \frac{2n^3+9n^2+13n-24}{24}.</math>
	În concluzie, pentru orice <math>n\ge 2 </math> are loc inegalitatea
	<math display="block"> \sum_{i=2}^{n} \sqrt[i]{\left(i!\right)^2} < \frac{2n^3+9n^2+13n-24}{24} .</math>

Andrei.Horvat: Created page with "'''23964 (Marin Bancoș)''' ''Să de demonstreze inegalitatea $\sum_{i=2}^{n} \sqrt[i]{\left(i!\right)^2} < \frac{2n^3+9n^2+13n-24}{24} .$ '' '''Soluție''' Pentru orice număr natural $n$ , cu $n\ge 2$ are loc inegalitatea $\sqrt[n]{n!} = \sqrt[n]{1\cdot 2\cdot \ldots \cdot n} < \frac{1+2+\ldots+n}{n} = \frac{n+1}{2}.$ Atunci $\sum_{i=2}^{n} \sqrt[i]{\left(I!\right)..."$

2025-09-17T17:50:42Z

Created page with "'''23964 (Marin Bancoș)''' ''Să de demonstreze inegalitatea <math display="block"> \sum_{i=2}^{n} \sqrt[i]{\left(i!\right)^2} < \frac{2n^3+9n^2+13n-24}{24} .</math>'' '''Soluție''' Pentru orice număr natural <math>n</math>, cu <math> n\ge 2</math> are loc inegalitatea <math display="block"> \sqrt[n]{n!} = \sqrt[n]{1\cdot 2\cdot \ldots \cdot n} < \frac{1+2+\ldots+n}{n} = \frac{n+1}{2}.</math> Atunci <math display="block"> \sum_{i=2}^{n} \sqrt[i]{\left(I!\right)..."

New page

'''[[23964]] (Marin Bancoș)'''

''Să de demonstreze inegalitatea <math display="block"> \sum_{i=2}^{n} \sqrt[i]{\left(i!\right)^2} < \frac{2n^3+9n^2+13n-24}{24} .</math>''

'''Soluție'''
Pentru orice număr natural <math>n</math>, cu <math> n\ge 2</math> are loc inegalitatea
<math display="block"> \sqrt[n]{n!} = \sqrt[n]{1\cdot 2\cdot \ldots \cdot n} < \frac{1+2+\ldots+n}{n} = \frac{n+1}{2}.</math>
Atunci
<math display="block"> \sum_{i=2}^{n} \sqrt[i]{\left(I!\right)2} < \sum_{i=2}^{n} \frac{\left(i+1\right)^2}{4}</math>
Suma <math>S_n=\sum_{i=2}^{n} \frac{\left(i+1\right)^2}{4}</math> poate fi calculată prin mai multe metode. De exemplu, avem
<math display="block">S_n = \frac{1}{4} \cdot \sum_{i=3}^{n+1} i^2 = \frac{1}{4} \left(\frac{n\left(n+1\right)\left(2n+1\right)}{6} - 5\right) = \frac{2n^3+9n^2+13n-24}{24}.</math>
În concluzie, pentru orice <math>n\ge 2 </math> are loc inegalitatea
<math display="block"> \sum_{i=2}^{n} \sqrt[i]{\left(i!\right)^2} < \frac{2n^3+9n^2+13n-24}{24} .</math>

23964 - Revision history

Andrei.Horvat at 17:56, 17 September 2025