27020

27020 (Gheorghe Szöllösy)

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

Soluție:

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

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

Avem $a_{n}=\left.{\frac {1}{2^{n}}}\right.C_{2n}^{n}$, 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 $\left.{\frac {(2n!)}{2^{n}(n!)^{3}}}\right..$