Evaluar la suma${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}$

Question

Evalúa la suma$${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.$ $

He intentado comparar esto con el problema similar aquí .

Creo que necesito diferenciar o integrar? Pero no estoy seguro de cómo podría funcionar eso.

¿Algunas ideas? Gracias.

Answer 1

Esto es$$\sum_{k=0}^{n}\frac{(-1)^{k}}{k+1}\binom{n}{k}.$$ Then multiplying each term by $ \ frac {n +1} {n +1},$ we get $$\sum_{k=0}^{n}\frac{(-1)^{k}}{n+1}\binom{n+1}{k+1}=\frac{1}{n+1}\sum_{k=1}^{n+1}(-1)^{k-1}\binom{n+1}{k}.$$ Adding and subtracting $ 1$ and applying the Binomial Theorem gives $$\frac{1}{n+1}-\frac{1}{n+1}\sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^{k}1^{n+1-k}=\frac{1}{n+1}(1-(1-1)^{n+1})=\frac{1}{n+1}.$PS

Answer 2

$$ \begin{align} S&=1-\frac 12\binom n1+\frac 13 \binom n2-\cdots+(-1)^n\frac 1{n+1}\binom nn\\ \times (n+1):\hspace{1cm}\\ (n+1)S&=(n+1)-\frac {n+1}2\binom n1+\frac {n+1}3\binom n2-\cdots+(-1)^n\frac {n+1}{n+1}\binom nn\\ &=\binom {n+1}1-\binom {n+1}2+\binom {n+1}3-\cdots +(-1)^n\binom {n+1}{n+1}\\ &=\color{blue}{\binom {n+1}0}\underbrace{\color{blue}{-\binom {n+1}0}+\binom {n+1}1-\binom {n+1}2+\binom {n+1}3-\cdots +(-1)^n\binom {n+1}{n+1}}_{=-\sum_{r=0}^{n+1}\binom {n+1}r(-1)^r=-(1-1)^{n+1}=0}\\ &=1\\ S&=\color{red}{\frac 1{n+1}} \end {align} $$

Answer 3

Insinuación

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