27020: Difference between revisions

Revision as of 17:17, 18 October 2023

27020 (Gheorghe Szöllösy)

Să se calculeze suma $\sum _{k=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }{\frac {1}{4^{k}\cdot (k!)^{2}\cdot (n-2k)!}},\quad n\geq 1$

Soluție:

Fie $a_{n}$ coeficientul lui $X^{n}$ din rezolvarea lui Failed to parse (syntax error): {\displaystyle P(X) = (X + \left[\dfrac{1}{2}\right])^2n = (X(1+X) + [\dfrac{1}{4}\right])^n = \sum_{k=0}^n C_n^k X^(n-k) \left[\dfrac{1}{4^k}\right]} .

Revision as of 17:16, 18 October 2023 edit Nagy Lenard (talk \| contribs) 188 edits No edit summary ← Older edit		Revision as of 17:17, 18 October 2023 edit undo Nagy Lenard (talk \| contribs) 188 edits No edit summary Newer edit →
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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> P(X) = (X + \left[\dfrac{1}{2}\right])^2n = (X(1+X) + [\dfrac{1}{4}\right])^n = \sum_{k=0}^n C_n^k X^(n-k) \left[\dfrac{1}{4^k}\right].		<math> P(X) = (X + \left[\dfrac{1}{2}\right])^2n = (X(1+X) + [\dfrac{1}{4}\right])^n = \sum_{k=0}^n C_n^k X^(n-k) \left[\dfrac{1}{4^k}\right]</math>.