Ta có 1/1x4+1/4x7+...+1/2002x2005
<=> =1/3.3(1/1x4+1/4x7+...+1/2002x2005)
=1/3(3/1x4+3/4x7+...+3/2002x2005)
=1/3(1-1/4+1/4-1/7+...+1/2002-1/2005)
=1/3(1-1/2005)
=1/3.2004/2005
=1.2004/3.2005
=668/2005
\(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+...+\(\frac{1}{2002.2005}\)=3(\(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+...+ \(\frac{1}{2002.2005}\)):3=(\(\frac{3}{1.4}\)+ \(\frac{3}{4.7}\)+...+ \(\frac{3}{2002.2005}\)):3= (1-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)):3=(1-\(\frac{1}{2005}\)) : 3 = \(\frac{668}{2005}\)
