27020: Difference between revisions
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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 | 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> | </math> | ||
'''Soluție:''' | |||
Fie a_n | |||
Revision as of 17:10, 18 October 2023
27020 (Gheorghe Szöllösy)
Să se calculeze suma
Soluție: Fie a_n