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 $a_{n}=\left({\frac {1}{2^{n}}}\right)C_{2}n^{n}$, iar pe de altă parte,

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

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

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