Vizualizare sursă pentru „27020”

'''27020 (Gheorghe Szöllösy)'''

''Să se calculeze suma'' <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2  (n-2k)!}, \quad n \geq 1.
</math>''

'''Soluție:'''

Fie <math> a_n </math>  coeficientul lui <math> X^n </math> din dezvoltarea lui

<math display="block"> P(X) = \left( X + \frac{1}{2}\right)^{2n} = \left(X(1+X) + \frac{1}{4}\right)^n = \sum_{k=0}^n C_n^k X^{n-k} (1+ 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,
<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}^2\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>

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