faktöriyel

Question

$\dfrac {1} {2!}+\dfrac {2} {3!}+\dfrac {3} {4!}+\ldots \dfrac {11} {12!}$ işleminin sonucu kaçtır?

Answer 1

İpucu=$\sum_{n=1}^{11} \frac{n}{(n+1)!}$

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

Answer 2

$\dfrac {1} {2!}+\dfrac {2} {3!}+\dfrac {3} {4!}+\ldots + \dfrac {11} {12!}=x$

$\dfrac {1} {2!}+\dfrac {2} {3!}+\dfrac {3} {4!}+\ldots + \dfrac {11} {12!}+$$(\dfrac {1} {2!}+\dfrac {1} {3!}+\dfrac {1} {4!}+\ldots + \dfrac {1} {12!})$$=$$x+$$(\dfrac {1} {2!}+\dfrac {1} {3!}+\dfrac {1} {4!}+\ldots + \dfrac {1} {12!})$

$1+\dfrac {1} {2!}+\dfrac {1} {3!}+\dfrac {1} {4!}+\ldots + \dfrac {1} {11!}=x+\dfrac {1} {2!}+\dfrac {1} {3!}+\dfrac {1} {4!}+\ldots + \dfrac {1} {12!}$

$x=1-\dfrac {1} {12!}$