Tübitak Lise 1. Aşama 1996 Soru 11

[matematikolimpiyati]

$\displaystyle \sum\limits_{n=1}^{9} \dfrac{3n+2}{n(n+1)(n+2)}$  toplamı aşağıdakilerden hangisine eşittir?

$\textbf{a)}\ \dfrac{293}{52}  \qquad\textbf{b)}\ \dfrac{189}{110}  \qquad\textbf{c)}\ \dfrac{179}{120}  \qquad\textbf{d)}\ \dfrac{3}{4}  \qquad\textbf{e)}\ \dfrac{5}{12}$

[Lokman Gökçe]

Yanıt: $\boxed{B}$

Verilen toplamı $$\begin{array}{lcl}
\displaystyle \sum\limits_{n=1}^{9} \dfrac{3n+2}{n(n+1)(n+2)} &=& \sum\limits_{n=1}^{9}\dfrac{3}{(n+1)(n+2)}  + \sum\limits_{n=1}^{9}\dfrac{2}{n(n+1)(n+2)} \\
&=& 3  \sum\limits_{n=1}^{9}\left(\dfrac{1}{n+1}- \dfrac{1}{n+2}\right)  + \sum\limits_{n=1}^{9}\left(\dfrac{1}{n(n+1)}- \dfrac{1}{(n+1)(n+2)}\right)
\end{array}$$ biçiminde teleskopik toplam olarak yazarsak, sadeleşmeler sonucunda $$3\left( \dfrac{1}{2} - \dfrac{1}{11}\right)  + \left( \dfrac{1}{2} - \dfrac{1}{110}\right) = \dfrac{189}{110}$$ elde ederiz.