14183: Difference between revisions

Revision as of 05:05, 9 January 2025

14183 (Gheorghe Szőllőssy)

Să se calculeze suma Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k} .

Soluție

Pentru orice număr natural Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} considerăm $S(p,n)=\displaystyle \sum _{k=0}^{n}k^{p}C_{n}^{k}$ .

Pentru orice număr natural Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} au loc egalitățileFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S(0,n) = \displaystyle \sum_{k=0}^n C_n^k = 2^n} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S(1,n) = \displaystyle \sum_{k=0}^n kC_n^k = n2^{n-1}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S(2,n) = \displaystyle \sum_{k=0}^n k^2C_n^k = n\left(n+1\right)2^{n-2}.}

Cum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 } , se obține Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S = n\left(n+1\right)2^{n-2} + 2\cdot n2^{n-1} + 2^n,} deciFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S = \displaystyle \sum_{k=0}^n \left(k+1\right)^2C_n^k = 2^{n-2}\left(n^2+5n+4\right).}

Observație

Pentru calculul sumei Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S(0,n)} se folosește dezvoltarea binomialăFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(X+1\right)^n = \displaystyle \sum_{k=0}^n C_n^k X^k.} Pentru calculul sumei Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S(1,n)} se folosește derivata obținută din dezvoltarea binomială Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(X+1\right)^n} , adică egalitateaFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n\left(X+1\right)^{n-1} = \displaystyle \sum_{k=1}^n kC_n^k X^{k-1}.} Pentru calculul sumei Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S(2,n)} , egalitatea precedentă se înmulțește cu Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X} , apoi se derivează. Se obțineFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}.}

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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>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(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>.
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