Tübitak Lise 1. Aşama 2003 Soru 02

[geo]

$1\cdot 2003 + 2\cdot 2002 + 3\cdot 2001 + \cdots + 2001 \cdot 3 + 2002 \cdot 2 + 2003 \cdot 1$ sayısının kaç asal böleni vardır?

$
\textbf{a)}\ 3
\qquad\textbf{b)}\ 4
\qquad\textbf{c)}\ 5
\qquad\textbf{d)}\ 6
\qquad\textbf{e)}\ 7
$

[geo]

Yanıt: $\boxed{C}$

$\begin{array}{lcl}
\sum\limits_{k=1}^{n} k \cdot (n+1-k) &=& (n+1)\sum\limits_{k=1}^{n} k - \sum\limits_{k=1}^{n} k^2 \\
&=& \dfrac{(n+1) \cdot n \cdot (n+1)}{2} - \dfrac{n \cdot (n+1) \cdot (2n+1)}{6} \\
&=& \dfrac{n(n+1)}{6} \cdot (3n+3 - 2n -1) \\
&=& \dfrac{n(n+1)(n+2)}{6} \\
\end{array}$

$n=2003$  için
$\sum\limits_{k=1}^{2003} k \cdot (2004-k) = \dfrac{2003 \cdot 2004 \cdot 2005}{6} = 334 \cdot 2003 \cdot 2005 = 2 \cdot 5 \cdot 167 \cdot 401 \cdot 2003$.