Question

Aflati valoare numarului natural nenul n pt care
$\frac{1}{1 \times 2} + \frac{2}{1 \times 2 \times 3} + \frac{3}{1 \times 2 \times 3 \times 4} + ... + \frac{n}{1 \times 2 \times 3 \times 4 \times ... \times n + 1} = \frac{(1 \times 2 \times 3 \times ... \times 2006) - 1}{1 \times 2 \times 3 \times ... \times 2006}$
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Answer 1

$\displaystyle \dfrac{1}{1\cdot 2}+\dfrac{2}{1\cdot 2\cdot 3}+\dfrac{3}{1\cdot 2\cdot 3\cdot 4}+...+\dfrac{n}{1\cdot 2\cdot 3\cdot 4\cdot...\cdot(n+1)} = \\ \\ = \sum\limits_{k=1}^n \dfrac{k}{(k+1)!} = \sum\limits_{k=1}^n\dfrac{k+1-1}{(k+1)!} = \sum\limits_{k=1}^n \dfrac{k+1}{(k+1)!} - \sum\limits_{k=1}^n\dfrac{1}{(k+1)!} =\\ \\ = \sum\limits_{k=1}^n \dfrac{k+1}{k!(k+1)}-\sum\limits_{k=1}^n\dfrac{1}{(k+1)!} = \sum\limits_{k=1}^n\dfrac{1}{k!} - \sum\limits_{k=1}^n\dfrac{1}{(k+1)!} =$

$= \dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3}!+...+\dfrac{1}{n!} - \dfrac{1}{2!}-\dfrac{1}{3!}-...-\dfrac{1}{n!}-\dfrac{1}{(n+1)!} = \\ \\ = \dfrac{1}{1!} - \dfrac{1}{(n+1)!} = \dfrac{(n+1)!-1}{(n+1)!}= \dfrac{\Big[1\cdot 2\cdot 3\cdot ...\cdot (n+1)\Big]-1}{1\cdot 2\cdot 3\cdot ...\cdot (n+1)} \\ \\ \\ \Rightarrow \boxed{n = 2005}$

Answer 2

Răspuns:

n=2005

Explicație pas cu pas:

$\frac{1}{2!} =\frac{2-1}{2!} =\frac{2}{2!} -\frac{1}{2!} =\frac{1}{1!} -\frac{1}{2!} \\\frac{2}{3!}=\frac{3-1}{3!}  =\frac{3}{3!} -\frac{1}{3!} =\frac{1}{2!} -\frac{1}{3!} \\....\\ \frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!} =\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!} -\frac{1}{(n+1)!}\\    Adunand aceste egalitati membru cu membru, obtinem in final:\\\frac{1}{2!} +\frac{2}{3!}+...+\frac{n}{(n+1)!}=1-\frac{1}{2!} +\frac{1}{2!}-\frac{1}{3!}+....+\frac{1}{n!}-\frac{1}{(n+1)!}$

$1-\frac{1}{(n+1)!}=\frac{2006!-1}{2006!}=>1-\frac{1}{(n+1)!}=1-\frac{1}{2006!}==>\\    =>(n+1)!=2006!=>n+1=2006=>n=2005$