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

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

Avem $a_{n}=\left({\frac {1}{2^{n}}}\right)C_{2}n^{n}$, 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. + ... = }

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

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