27020: Difference between revisions

Revision as of 17:08, 18 October 2023

27020 (Gheorghe Szöllösy)

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

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 '''27020 (Gheorghe Szöllösy)'''
-Să se calculeze suma <math> \sum_{k=0}^\left[\frac{n}{2}\right]  \[left\dfrac{1}{4^k \cdot (k!)^2(n-2k)!}\right], n \geq 1 </math>
+Să se calculeze suma <math> \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} \frac{1}{4^k \cdot (k!)^2 \cdot (n-2k)!}, \quad n \geq 1
+</math>