Question

calcular S=(2n 0)+(2n 1)+(2n 2)+(2n 3)+...+(2n n-1)+[2^(n-1)].(2n-1)!! /n!! donde n∈Z⁺

Answer 1

algunas propiedades:

$(2n-1)!!=\dfrac{(2n)!}{2^{n}n!}$

$C^{n}_{k}=C^{n}_{n-k}$ complemento

$C^{n}_{0}+C^{n}_{1}+C^{n}_{2}+C^{n}_{3}+...+C^{n}_{n-1}+C^{n}_{n}=2^{n}$

analizando

$S=\dbinom{2n}{0}+\dbinom{2n}{1}+\dbinom{2n}{2}+\dbinom{2n}{3}+...+\dbinom{2n}{n-1}+\dfrac{2^{n-1}(2n-1)!!}{n!}$

analizamos la ultima parte

$\dfrac{2^{n-1}(2n-1)!!}{n!}=\dfrac{2^{n-1}}{n!}.(2n-1)!!$

$\dfrac{2^{n-1}}{n!}.\dfrac{(2n)!}{2^{n}(n)!}=\dfrac{1}{2}.\dfrac{(2n)!}{n!.n!}=\dfrac{1}{2}C^{2n}_{n}$

remplazando

$S=\dbinom{2n}{0}+\dbinom{2n}{1}+\dbinom{2n}{2}+\dbinom{2n}{3}+...+\dbinom{2n}{n-1}+\dfrac{1}{2}C^{2n}_{n}$

$S=C^{2n}_{0}+C^{2n}_{1}+C^{2n}_{2}+C^{2n}_{3}+...+C^{2n}_{n-2}+C^{2n}_{n-1}+\dfrac{1}{2}C^{2n}_{n}$

hallemos una la expresión ,pero del complemento de cada termino

$S=C^{2n}_{2n}+C^{2n}_{2n-1}+C^{2n}_{2n-2}+C^{2n}_{2n-3}+...+C^{2n}_{n+2}+C^{2n}_{n+1}+\dfrac{1}{2}C^{2n}_{n}$

sumamos la dos expresiones

$2S=[C^{2n}_{0}+C^{2n}_{1}+C^{2n}_{2}+...+C^{2n}_{n-1}+\dfrac{1}{2}C^{2n}_{n}]+[\dfrac{1}{2}C^{2n}_{n}+C^{2n}_{n+1}+C^{2n}_{n+2}+..+C^{2n}_{2n}]$

$2S=[C^{2n}_{0}+C^{2n}_{1}+C^{2n}_{2}+...+C^{2n}_{n-1}+C^{2n}_{n}+C^{2n}_{n+1}+C^{2n}_{n+2}+..+C^{2n}_{2n}]$

$2S=2^{2n}$

$S=2^{2n-1}$

$\mathbb{AUTOR} \ SmithValdez$

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