27020

27020 (Gheorghe Szöllösy)

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

Soluție:

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

$$P(X)=\left(X+\left\lfloor {\frac {1}{2}}\right\rfloor \right)^{2n}=\left(X(1+X)+\left\lfloor {\frac {1}{4}}\right\rfloor \right)^{n}=\sum _{k=0}^{n}C_{n}^{k}X^{(n-k)}\left({\frac {1}{4^{k}}}\right).$$

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, 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 = 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. + ... = }

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

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