14183 - Revision history

Andrei.Horvat at 05:06, 9 January 2025

2025-01-09T05:06:54Z

← Older revision		Revision as of 05:06, 9 January 2025
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	Pentru orice număr natural <math>n</math> au loc egalitățile<math display="block">S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n</math><math display="block">S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}</math><math display="block">S(2,n) = \displaystyle \sum_{k=0}^n k^2C_n^k = n\left(n+1\right)2^{n-2}.</math>		Pentru orice număr natural <math>n</math> au loc egalitățile<math display="block">S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n</math><math display="block">S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}</math><math display="block">S(2,n) = \displaystyle \sum_{k=0}^n k^2C_n^k = n\left(n+1\right)2^{n-2}.</math>


	Cum <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = \displaystyle \sum_{k=0}^n k^2C_n^k + 2\cdot \displaystyle \sum_{k=0}^n kC_n^k + \displaystyle \sum_{k=0}^n C_n^k </math>, se obține <math display="block"> S = n\left(n+1\right)2^{n-2} + 2\cdot n2^{n-1} + 2^n,</math>deci<math display="block">S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = 2^{n-2}\left(n^2+5n+4\right).</math>		Cum <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = \displaystyle \sum_{k=0}^n k^2C_n^k + 2\cdot \displaystyle \sum_{k=0}^n kC_n^k + \displaystyle \sum_{k=0}^n C_n^k </math>, se obține <math display="block"> S = n\left(n+1\right)2^{n-2} + 2\cdot n2^{n-1} + 2^n,</math>deci<math display="block">S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = 2^{n-2}\left(n^2+5n+4\right).</math>

Andrei.Horvat at 05:05, 9 January 2025

2025-01-09T05:05:38Z

← Older revision		Revision as of 05:05, 9 January 2025
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	Pentru orice număr natural <math>p</math> considerăm <math>S(p,n) = \displaystyle \sum_{k=0}^n k^pC_n^k</math>.		Pentru orice număr natural <math>p</math> considerăm <math>S(p,n) = \displaystyle \sum_{k=0}^n k^pC_n^k</math>.

	Pentru orice număr natural <math>n</math> au loc egalitățile<math display="block">S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n</math><math display="block">S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}</math><math display="block">S(2,n) = \displaystyle \sum_{k=0}^n k^2C_n^k = n\left(n+1\right)2^{n-2}</math>Cum <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = \displaystyle \sum_{k=0}^n k^2C_n^k + 2\cdot \displaystyle \sum_{k=0}^n kC_n^k + \displaystyle \sum_{k=0}^n C_n^k </math>, se obține <math display="block"> S = n\left(n+1\right)2^{n-2} + 2\cdot n2^{n-1} + 2^n,</math>deci <math display="block">S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = 2^{n-2}\left(n^2+5n+4\right).</math>		Pentru orice număr natural <math>n</math> au loc egalitățile<math display="block">S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n</math><math display="block">S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}</math><math display="block">S(2,n) = \displaystyle \sum_{k=0}^n k^2C_n^k = n\left(n+1\right)2^{n-2}.</math>


			Cum <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = \displaystyle \sum_{k=0}^n k^2C_n^k + 2\cdot \displaystyle \sum_{k=0}^n kC_n^k + \displaystyle \sum_{k=0}^n C_n^k </math>, se obține <math display="block"> S = n\left(n+1\right)2^{n-2} + 2\cdot n2^{n-1} + 2^n,</math>deci<math display="block">S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = 2^{n-2}\left(n^2+5n+4\right).</math>

	'''Observație'''		'''Observație'''

	Pentru calculul sumei <math>S(0,n)</math> se folosește dezvoltarea binomială <math display="block">\left(X+1\right)^n = \displaystyle \sum_{k=0}^n C_n^k X^k</math>.		Pentru calculul sumei <math>S(0,n)</math> se folosește dezvoltarea binomială<math display="block">\left(X+1\right)^n = \displaystyle \sum_{k=0}^n C_n^k X^k.</math>Pentru calculul sumei <math>S(1,n)</math> se folosește derivata obținută din dezvoltarea binomială <math>\left(X+1\right)^n</math>, adică egalitatea<math display="block">n\left(X+1\right)^{n-1} = \displaystyle \sum_{k=1}^n kC_n^k X^{k-1}.</math>Pentru calculul sumei <math>S(2,n)</math>, egalitatea precedentă se înmulțește cu <math>X</math>, apoi se derivează. Se obține<math display="block">n\left(X+1\right)^{n-1} + n\left(n-1\right)X\left(X+1\right)^{n-2} = \displaystyle \sum_{k=1}^n k^2C_n^k X^{k-1}.</math>

	Pentru calculul sumei <math>S(1,n)</math> se folosește derivata obținută din dezvoltarea binomială <math>\left(X+1\right)^n</math>, adică egalitatea <math display="block">n\left(X+1\right)^{n-1} = \displaystyle \sum_{k=1}^n kC_n^k X^{k-1}</math>.

	Pentru calculul sumei <math>S(2,n)</math>, egalitatea precedentă se înmulțește cu <math>X</math>, apoi se derivează. Se obține <math display="block">n\left(X+1\right)^{n-1} + n\left(n-1\right)X\left(X+1\right)^{n-2} = \displaystyle \sum_{k=1}^n k^2C_n^k X^{k-1}</math>.		.

Andrei.Horvat at 05:04, 9 January 2025

2025-01-09T05:04:02Z

← Older revision		Revision as of 05:04, 9 January 2025
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	Pentru orice număr natural <math>n</math> au loc egalitățile<math display="block">S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n</math><math display="block">S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}</math><math display="block">S(2,n) = \displaystyle \sum_{k=0}^n k^2C_n^k = n\left(n+1\right)2^{n-2}</math>Cum <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = \displaystyle \sum_{k=0}^n k^2C_n^k + 2\cdot \displaystyle \sum_{k=0}^n kC_n^k + \displaystyle \sum_{k=0}^n C_n^k </math>, se obține <math display="block"> S = n\left(n+1\right)2^{n-2} + 2\cdot n2^{n-1} + 2^n,</math>deci <math display="block">S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = 2^{n-2}\left(n^2+5n+4\right).</math>		Pentru orice număr natural <math>n</math> au loc egalitățile<math display="block">S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n</math><math display="block">S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}</math><math display="block">S(2,n) = \displaystyle \sum_{k=0}^n k^2C_n^k = n\left(n+1\right)2^{n-2}</math>Cum <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = \displaystyle \sum_{k=0}^n k^2C_n^k + 2\cdot \displaystyle \sum_{k=0}^n kC_n^k + \displaystyle \sum_{k=0}^n C_n^k </math>, se obține <math display="block"> S = n\left(n+1\right)2^{n-2} + 2\cdot n2^{n-1} + 2^n,</math>deci <math display="block">S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = 2^{n-2}\left(n^2+5n+4\right).</math>

			'''Observație'''

			Pentru calculul sumei <math>S(0,n)</math> se folosește dezvoltarea binomială <math display="block">\left(X+1\right)^n = \displaystyle \sum_{k=0}^n C_n^k X^k</math>.

			Pentru calculul sumei <math>S(1,n)</math> se folosește derivata obținută din dezvoltarea binomială <math>\left(X+1\right)^n</math>, adică egalitatea <math display="block">n\left(X+1\right)^{n-1} = \displaystyle \sum_{k=1}^n kC_n^k X^{k-1}</math>.

			Pentru calculul sumei <math>S(2,n)</math>, egalitatea precedentă se înmulțește cu <math>X</math>, apoi se derivează. Se obține <math display="block">n\left(X+1\right)^{n-1} + n\left(n-1\right)X\left(X+1\right)^{n-2} = \displaystyle \sum_{k=1}^n k^2C_n^k X^{k-1}</math>.

Andrei.Horvat at 13:32, 8 January 2025

2025-01-08T13:32:19Z

← Older revision		Revision as of 13:32, 8 January 2025
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	'''Soluție'''		'''Soluție'''

	Pentru orice număr natural <math>p</math> considerăm <math>S(p,n) = \displaystyle \sum_{k=0}^n k^pC_n^k</math>.		Pentru orice număr natural <math>p</math> considerăm <math>S(p,n) = \displaystyle \sum_{k=0}^n k^pC_n^k</math>.
	~~Pentru orice număr natural <math>n</math> au loc egalitățile~~

	~~<math>S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n</math>~~

	~~<math>S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}</math>~~

	~~<math>S(2,n) = \displaystyle \sum_{k=0}^n k^2C_n^k = n\left(n+1\right)2^{n-2}</math>.~~

	Cum <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = \displaystyle \sum_{k=0}^n k^2C_n^k + 2\cdot \displaystyle \sum_{k=0}^n kC_n^k + \displaystyle \sum_{k=0}^n C_n^k </math>, se obține <math> S = n\left(n+1\right)2^{n-2} + 2\cdot n2^{n-1} + 2^n</math>, deci <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = 2^{n-2}\left(n^2+5n+4\right)</math>.		Pentru orice număr natural <math>n</math> au loc egalitățile<math display="block">S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n</math><math display="block">S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}</math><math display="block">S(2,n) = \displaystyle \sum_{k=0}^n k^2C_n^k = n\left(n+1\right)2^{n-2}</math>Cum <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = \displaystyle \sum_{k=0}^n k^2C_n^k + 2\cdot \displaystyle \sum_{k=0}^n kC_n^k + \displaystyle \sum_{k=0}^n C_n^k </math>, se obține <math display="block"> S = n\left(n+1\right)2^{n-2} + 2\cdot n2^{n-1} + 2^n,</math>deci <math display="block">S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = 2^{n-2}\left(n^2+5n+4\right).</math>

Andrei.Horvat: Created page with "'''14183 (Gheorghe Szőllőssy)''' ''Să se calculeze suma $S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k$ .'' '''Soluție''' Pentru orice număr natural $p$ considerăm $S(p,n) = \displaystyle \sum_{k=0}^n k^pC_n^k$ . Pentru orice număr natural $n$ au loc egalitățile $S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n$ $S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}$ $S(2,n) = \dis..."$

2025-01-08T13:30:36Z

Created page with "'''14183 (Gheorghe Szőllőssy)''' ''Să se calculeze suma <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k</math>.'' '''Soluție''' Pentru orice număr natural <math>p</math> considerăm <math>S(p,n) = \displaystyle \sum_{k=0}^n k^pC_n^k</math>. Pentru orice număr natural <math>n</math> au loc egalitățile <math>S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n</math> <math>S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}</math> <math>S(2,n) = \dis..."

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'''14183 (Gheorghe Szőllőssy)'''

''Să se calculeze suma <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k</math>.''

'''Soluție'''
Pentru orice număr natural <math>p</math> considerăm <math>S(p,n) = \displaystyle \sum_{k=0}^n k^pC_n^k</math>.
Pentru orice număr natural <math>n</math> au loc egalitățile

<math>S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n</math>

<math>S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}</math>

<math>S(2,n) = \displaystyle \sum_{k=0}^n k^2C_n^k = n\left(n+1\right)2^{n-2}</math>.

Cum <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = \displaystyle \sum_{k=0}^n k^2C_n^k + 2\cdot \displaystyle \sum_{k=0}^n kC_n^k + \displaystyle \sum_{k=0}^n C_n^k </math>, se obține <math> S = n\left(n+1\right)2^{n-2} + 2\cdot n2^{n-1} + 2^n</math>, deci <math>S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = 2^{n-2}\left(n^2+5n+4\right)</math>.