$S = \dfrac{n \choose 0} {1} + \dfrac{n \choose 1} {2} + \dfrac{n \choose 2} {3} + \dotsb + \dfrac {n \choose n} {n+1} $
Respuestas
¿Demasiados anuncios?
Marco Cantarini
Puntos
10794
Tenemos $$ \sum_ {k = 0} ^ {n} \dbinom {n} {k} x ^ {k} = \left (1 + x\right) ^ {n} $$ then $$\int_{0}^{1}\left (1 + x\right) ^ {n} dx = \sum_ {k = 0} ^ {n} \dbinom {n} {k} \int_ {0} ^ {1} x ^ {k} dx = \sum_ {k = 0} ^ {n} \dbinom {n} {k} \frac {1} {k+1} $$ and so $ $\sum_{k=0}^{n}\dbinom{n} {} k} \frac {1} {k+1} = \int_ {1} ^ {2} u ^ du {n} = \frac {2 ^ {n + 1} -1} {n+1}. $$
$\bf{My\; Solution::}$ $$\displaystyle S = \sum_{k=0}^{n}\binom{n}{k}\cdot \frac{1}{k+1} = \sum_{k=0}^{n}\frac{n!}{k!\cdot (n-k)!}\cdot \frac{1}{k+1}$ $ Podemos escribir
$$\displaystyle S = \frac{1}{n+1}\sum_{k=0}^{n}\frac{(n+1)!}{(k+1)!\cdot (n-k)!}=\frac{1}{n+1}\sum_{k=0}^{n}\binom{n+1}{k+1} = \frac{2^{n+1}-1}{n+1}.$$
Arriba hemos usado la identidad $$\displaystyle \sum_{k=0}^{n}\binom{n}{k} = 2^n$ $.
Para %#% $ #%
