$$\begin{align*} & \forall n\in \mathbb{N} \\ & \sum _{k=1}^{n}\dfrac {k} {\left( k+1\right) !}=1-\dfrac {1} {\left( n+1\right) !}\end{align*}$$ Olduğunu tümevarım yöntemiyle gösteriniz.

Question

$$\begin{align*} & \forall n\in \mathbb{N} \\ & \sum _{k=1}^{n}\dfrac {k} {\left( k+1\right) !}=1-\dfrac {1} {\left( n+1\right) !}\end{align*}$$  

Answer 1

$n=1$ icin esitlik dogru.

$n\geq1$ icin dogru ise $n+1$ icin de dogru:

$$\sum\limits_{k=0}^{n+1}\frac{k}{(k+1)!}=\sum\limits_{k=0}^{n}\frac{k}{(k+1)!}+\frac{n+1}{(n+2)!}$$ $$=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}=1-\frac{1}{(n+2)!}.$$

Comments

Sağolasın ama anlayamadım yapmak istedğini

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