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Linia 14: |
Linia 14: |
| + C_n^2 \cdot C_(n-2)^1\left.\frac{1}{4^2}\right. + ... = </math> | | + C_n^2 \cdot C_(n-2)^1\left.\frac{1}{4^2}\right. + ... = </math> |
| <math display="block"> = \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} C_n^k C_(n-k)^k \cdot \left.\frac{1}{4^k}\right. = n! \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} \frac{1}{(k!)^2 (n-k)! 4^k},</math> | | <math display="block"> = \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} C_n^k C_(n-k)^k \cdot \left.\frac{1}{4^k}\right. = n! \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} \frac{1}{(k!)^2 (n-k)! 4^k},</math> |
| | |
| | deci suma este egală cu <math> \left.\frac{(2n!}{2^n(n!)^3}\right. .</math> |
Versiunea de la data 18 octombrie 2023 17:36
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.+...=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4a1fc68738185155e6b10629b1addd74391000a)
![{\displaystyle =\sum _{k=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }C_{n}^{k}C_{(}n-k)^{k}\cdot \left.{\frac {1}{4^{k}}}\right.=n!\sum _{k=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }{\frac {1}{(k!)^{2}(n-k)!4^{k}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c73e099188f38ae6dbe7ae08c9bc79d26b256c9)
deci suma este egală cu