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| '''27020 (Gheorghe Szöllösy)''' | | '''27020 (Gheorghe Szöllösy)''' |
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| ''Să se calculeze suma'' <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1 | | ''Să se calculeze suma'' <math> \sum_{k=0}^{\left[\frac{n}{2}\right]} \frac{1}{4^k \cdot (k!)^2 (n-2k)!}, \quad n \geq 1. |
| </math>.'' | | </math>'' |
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| '''Soluție:''' | | '''Soluție:''' |
Versiunea de la data 18 octombrie 2023 18:06
27020 (Gheorghe Szöllösy)
Să se calculeze suma
Soluție:
Fie
coeficientul lui
din rezolvarea lui
![{\displaystyle P(X)=\left(X+{\frac {1}{2}}\right)^{2n}=\left(X(1+X)+{\frac {1}{4}}\right)^{n}=\sum _{k=0}^{n}C_{n}^{k}X^{n-k}(!-X)^{n-k}\left.{\frac {1}{4^{k}}}\right..}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c4951e3a38773bf54e3dd50d824e8bbac571972)
Avem
, iar pe de altă parte,
![{\displaystyle 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.+...=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e71472f212e44ab4a5e0471bce42260754c3f6ab)
![{\displaystyle =\sum _{k=0}^{\left[{\frac {n}{2}}\right]}C_{n}^{k}C_{n-k}^{k}\cdot \left.{\frac {1}{4^{k}}}\right.=n!\sum _{k=0}^{\left[{\frac {n}{2}}\right]}{\frac {1}{(k!)^{2}(n-k)!4^{k}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90ffe8287db2b247fed3fdc3c7941add52185dc2)
deci suma este egală cu