27020: Difference between revisions

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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) + C_n^2 \cdot C_(n-2)^1\left(\frac{1}{4^2}\right) +  ... = </math>
<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>
<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 17:33, 18 October 2023

27020 (Gheorghe Szöllösy)

Să se calculeze suma

Soluție:

Fie coeficientul lui din rezolvarea lui

Avem , iar pe de altă parte, Failed to parse (syntax error): {\displaystyle 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) + ... = } Failed to parse (syntax error): {\displaystyle = \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},}