27020

27020 (Gheorghe Szöllösy)

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

Soluție:

Fie $a_{n}$ coeficientul lui $X^{n}$ din rezolvarea lui

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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)}

Avem Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n = \left(\frac{1}{2^n}\right) C_2n^n } , iar pe de altă parte,

$$a_{n}=C_{n}^{0}\cdot C_{n}^{0}+C_{n}^{1}\cdot C_{n-1}^{1}\left.{\frac {1}{4}}\right.+C_{n}^{2}\cdot C_{n-2}^{1}\left.{\frac {1}{4^{2}}}\right.+...=$$

$$=\sum _{k=0}^{\left[{\frac {n}{2}}\right]}C_{n}^{k}C_{(}n-k)^{k}\cdot \left.{\frac {1}{4^{k}}}\right.=n!\sum _{k=0}^{\left[{\frac {n}{2}}\right]}{\frac {1}{(k!)^{2}(n-k)!4^{k}}},$$

deci suma este egală cu $\left.{\frac {(2n!}{2^{n}(n!)^{3}}}\right..$