27020: Difference between revisions

From Bitnami MediaWiki
No edit summary
No edit summary
Line 13: Line 13:
<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\lfloor\frac{n}{2}\right\rfloor} C_n^k C_(n-k)^k \cdot \left(\frac{1}{4^k}\right) = n! \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} \frac{1}{(k!)^2 (n-k)! 4^k},</math>
<math display="block"> = \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} C_n^k C_(n-k)^k \cdot \left.\frac{1}{4^k}\right. = n! \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} \frac{1}{(k!)^2 (n-k)! 4^k},</math>

Revision as of 17:34, 18 October 2023

27020 (Gheorghe Szöllösy)

Să se calculeze suma

Soluție:

Fie coeficientul lui din rezolvarea lui

Avem , iar pe de altă parte,