27020 (Gheorghe Szöllösy)
Să se calculeze suma ∑ k = 0 [ n 2 ] 1 4 k ⋅ ( k ! ) 2 ( n − 2 k ) ! , n ≥ 1. {\displaystyle \sum _{k=0}^{\left[{\frac {n}{2}}\right]}{\frac {1}{4^{k}\cdot (k!)^{2}(n-2k)!}},\quad n\geq 1.}
![{\displaystyle \sum _{k=0}^{\left[{\frac {n}{2}}\right]}{\frac {1}{4^{k}\cdot (k!)^{2}(n-2k)!}},\quad n\geq 1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1963241c69cc0a80c84902a9ace61e99e7bfda6)