Tübitak Lise 1.Aşama 1995 Soru 08

[alpercay]

$\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\cdots +\dfrac{99}{100!}$ toplamı neye eşittir?

$\textbf{a)}\ 1+\dfrac{99}{100!} \qquad\textbf{b)}\ \dfrac{101}{100} \qquad\textbf{c)}\ 1-\dfrac{99}{100} \qquad\textbf{d)}\ 1 \qquad\textbf{e)}\ 1-\dfrac{1}{100!}$

[Lokman Gökçe]

Yanıt: $\boxed{E}$

$\displaystyle{ S=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!} = \sum_{n=1}^{99}\dfrac{n}{(n+1)!} }$ yazabiliriz. Aşağıdaki teleskopik toplamı oluşturarak,
$\displaystyle{ S=  \sum_{n=1}^{99}\dfrac{(n+1)-1}{(n+1)!} =  \sum_{n=1}^{99} \left( \dfrac{n+1}{(n+1)!} -  \dfrac{1}{(n+1)!} \right)=  \sum_{n=1}^{99} \left(\dfrac{1}{n!} -  \dfrac{1}{(n+1)!} \right) } = 1 - \dfrac{1}{100!} $ elde ederiz.