27020: Difference between revisions
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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 + \left\lfloor\frac{1}{2}\right\rfloor\right)^{2n} = \left(X(1+X) + \left\lfloor\frac{1}{4}\right\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 + \left\lfloor\frac{1}{2}\right\rfloor\right)^{2n} = \left(X(1+X) + \left\lfloor\frac{1}{4}\right\rfloor\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, | ||
Revision as of 17:23, 18 October 2023
27020 (Gheorghe Szöllösy)
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 \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} \frac{1}{4^k \cdot (k!)^2 \cdot (n-2k)!}, \quad n \geq 1 }
Soluție:
Fie 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 a_n } coeficientul lui din rezolvarea lui
Avem 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 a_n = \left(\frac{1}{2^n}\right) C_2n^n } , iar pe de altă parte,