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/7d6a389d7e2b3f110ed04c5541d3fdb8a9691c8f)