27020: Difference between revisions

From Bitnami MediaWiki
No edit summary
No edit summary
Line 8: Line 8:
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 + \frac{1}{2}\right)^{2n} = \left(X(1+X) + \lfloor 1/4 \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 + \frac{1}{2}\right)^{2n} = (X(1+X) + \left.\frac{1}{4}\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,
<math display="block"> a_n = C_n^0 \cdot C_n^0 + C_n^1 \cdot C_(n-1)^1 \left.\frac{1}{4}\right.
<math display="block"> a_n = C_n^0 \cdot C_n^0 + C_n^1 \cdot C_{n-1}^1 \left.\frac{1}{4}\right.
  + C_n^2 \cdot C_(n-2)^1\left.\frac{1}{4^2}\right. +  ... = </math>
  + C_n^2 \cdot C_{n-2}^1\left.\frac{1}{4^2}\right. +  ... = </math>
<math display="block"> = \sum_{k=0}^{\left[\frac{n}{2}\right]} C_n^k C_(n-k)^k \cdot \left.\frac{1}{4^k}\right. = n! \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{(k!)^2 (n-k)! 4^k},</math>
<math display="block"> = \sum_{k=0}^{\left[\frac{n}{2}\right]} C_n^k C_(n-k)^k \cdot \left.\frac{1}{4^k}\right. = n! \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{(k!)^2 (n-k)! 4^k},</math>


deci suma este egală cu <math> \left.\frac{(2n!}{2^n(n!)^3}\right. .</math>
deci suma este egală cu <math> \left.\frac{(2n!}{2^n(n!)^3}\right. .</math>

Revision as of 17:48, 18 October 2023

27020 (Gheorghe Szöllösy)

Să se calculeze suma

Soluție:

Fie coeficientul lui din rezolvarea lui

Avem , iar pe de altă parte,

deci suma este egală cu