ข้อ 8 นะครับ
$(1-\frac{1}{2})(1-\frac{1}{3})....(1-\frac{1}{2002})(1-\frac{1}{2003})(1+\frac{1}{2})(1+\frac{1}{3})....(1+\frac{1}{2002})(1+\frac{1}{2003})$
$=(\frac{1}{2})(\frac{2}{3}).....(\frac{2001}{2002})(\frac{2002}{2003})(\frac{3}{2})(\frac{4}{3}).....(\frac{2003}{2002})(\frac{ 2004}{2003})$
$=(\frac{1}{2003})(\frac{2004}{2})$
$=\frac{1002}{2003}$
ถูกผิดก็ติเตียนกันด้วยนะครับ
|