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Linia 11: |
Linia 11: |
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| Avem <math> a_n = \left(\frac{1}{2^n}\right) C_2n^n </math>, iar pe de altă parte, | | Avem <math> a_n = \left(\frac{1}{2^n}\right) C_2n^n </math>, iar pe de altă parte, |
| <math display="block"> 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) + ... = \\ | | <math display="block"> 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) + ... = |
| = \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} C_n^k C_(n-k)^k \cdot \left(\frac{1}{4^k}\right) </math> | | = \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} C_n^k C_(n-k)^k \cdot \left(\frac{1}{4^k}\right) </math> |
Versiunea de la data 18 octombrie 2023 17:29
27020 (Gheorghe Szöllösy)
Să se calculeze suma
Soluție:
Fie
coeficientul lui
din rezolvarea lui
![{\displaystyle P(X)=\left(X+\left\lfloor {\frac {1}{2}}\right\rfloor \right)^{2n}=\left(X(1+X)+\left\lfloor {\frac {1}{4}}\right\rfloor \right)^{n}=\sum _{k=0}^{n}C_{n}^{k}X^{(n-k)}\left({\frac {1}{4^{k}}}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b21035320eca64ad4a7cde5087c47d709b2228dc)
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)+...==\sum _{k=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }C_{n}^{k}C_{(}n-k)^{k}\cdot \left({\frac {1}{4^{k}}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/196cc857d54784ab360a7c41992534e2ba62dcc2)