Computación

Question

$$\sum_{k=0}^{100}\frac{1}{(100-k)!(100+k)!}$$

Mi trabajo $$\sum_{k=0}^{100}\frac{2n!}{(2n!)(n-k)!(n+k)!}$ $

$$\sum_{k=0}^{100}\frac{^{2n}C_{n-k}}{(2n!)}$ $ $$\sum_{k=0}^{100}\frac{^{2n}C_{n+k}}{(2n!)}$ $ ¿Cómo debo seguir adelante?

Answer 1

Tratando de analizar su solicitud, queremos calcular ($n=100$, o quizás de cualquiera $n\in\mathbb{N}$) %#% $ #% debido a la simetría de los coeficientes binomiales, $$ \sum_{k=0}^{n}\frac{1}{(n-k)!(n+k)!} = \sum_{k=0}^{n}\frac{1}{k!(2n-k)!} = [x^{2n}]\left(\sum_{k=0}^{n}\frac{x^k}{k!}\right)^2=\frac{1}{(2n)!}\sum_{k=0}^{n}\binom{2n}{k}. $ $ por lo tanto:

$$ \sum_{k=0}^{n}\binom{2n}{k} = \frac{1}{2}\left( 4^n+\binom{2n}{n}\right),$$

Answer 2

Probablemente demasiado compleja para una respuesta.

Teniendo en cuenta que $$\sum_{k=0}^{n}\frac{x^k}{(n-k)!(n+k)!}=\frac{\, _2F_1(1,-n;n+1;-x)}{(n!)^2}$$ Setting $ x=1$, then $$\sum_{k=0}^{n}\frac{1}{(n-k)! () n + k)!} = \frac{\Gamma \left(n+\frac{1}{2}\right) + \sqrt {\pi} \,\Gamma (n + 1)} {2 (n)! ^ \Gamma \left(n+\frac{1}{2}\right) 2} = \frac {1} {2} \left (\frac {1} {\Gamma (n + 1) ^ 2} + \frac {4 ^ n} {\Gamma (2 n + 1)} \right) = \frac {1} {2} \left(\frac{1}{(n!) ^ 2} + \frac {4 ^ n} {(2 n)!} \right)$$ Using Stirling approximation for $k!$, an approximation could be $$\frac{e^{2 n} \left(\sqrt{\pi n} +1\right)}{4 \pi n^{(2 n+1)} }$$

Answer 3

Solucionar el problema de $n \in \mathbb{N}$ en vez de solo el $n = 100$, obtendrá\begin{align} \sum_{k=0}^{n} \frac{1}{(n-k)!(n+k)!} &= \frac{1}{(2n)!} \sum_{k=n}^{2n} \frac{(2n)!}{k!(2n-k)!} = \frac{1}{(2n)!} \sum_{k=n}^{2n} \binom{2n}{k} \\ &= \frac{1}{2(2n)!} \left( \binom{2n}{n} + 2^{2n} \right) \\ \end {Alinee el} en el último paso se utiliza el hecho de que \sum_{k=n}^{2n $$} \binom{2n}{k} = \frac{1}{2} \left (\binom{2n}{n} + 2 ^ {2n} \right) $$ que se muestra por\begin{align} 2^{2n} &= \sum_{k=0}^{2n} \binom{2n}{k} \\ &= \sum_{k=0}^{n-1} \binom{2n}{k} ~+~ \binom{2n}{n} ~+~ \sum_{k=n+1}^{2n} \binom{2n}{k} \\ &= \sum_{k=n+1}^{2n} \binom{2n}{2n-k} ~+~ \binom{2n}{n} ~+~ \sum_{k=n+1}^{2n} \binom{2n}{k} \\ &= \sum_{k=n+1}^{2n} \binom{2n}{k} ~+~ \binom{2n}{n} ~+~ \sum_{k=n+1}^{2n} \binom{2n}{k} \\ &= 2 \sum_{k=n}^{2n} \binom{2n}{k} ~-~ \binom{2n}{n} \end {Alinee el} de que tienes $$ \frac{1}{2} \left (2 ^ {2n} + \binom{2n}{n} \right) = \sum_{k=n}^{2n} \binom{2n}{2n-k} $$