a)Tính M=1/3 + 1/6 + 1/10 + 1/15 +……+ 1/2004.2005 b)Tính S= 1/1x2x3 + 1/2x3x4 + 1/3x4x5 + …..+ 1/98.99.100

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1. Giải thích các bước giải:

   $M = \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + … + \frac{1}{2004.2005}$

   ⇔$\frac{1}{2}M = \frac{1}{6} + \frac{1}{12} + … + \frac{1}{2004.2005}$

   ⇔$\frac{1}{2}M = \frac{1}{2.3} + \frac{1}{3.4} + … + \frac{1}{2004.2005}$

   ⇔$\frac{1}{2}M = \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} +  … + \frac{1}{2004} – \frac{1}{2005}$

   ⇔$\frac{1}{2}M = \frac{1}{2} – \frac{1}{2005}$

   ⇔$\frac{1}{2}M = \frac{2003}{4010}$

   ⇔$M = \frac{2003}{4010} : \frac{1}{2}$

   ⇔$M = \frac{2003}{2005}$

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2. Đáp án:

    a)2003/2005

   b)4949/9900

   Giải thích các bước giải:

    a)M=1/3+1/6+1/10+1/15+….+1/2004.2005

   M.1/2=1/6+1/12+1/20+1/30+….+1/2004.2005

   M.1/2=1/2.3=1/3.4+1/4.5+1/5.6+….+1/2004.2005

   M.1/2=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+….+1/2004-1/2005

   M.1/2=1/2-1/2005

   M.1/2=2003/4010

   =>M=2003/2005

   b)S=1/1.2.3+1/2.3.4+1/3.4.5+….+1/98.99.100

   =2S=2x(1/1.2.3+1/2.3.4+1/3.4.5+…+1/98.99.100)

   =2/1.2.3-2/2.3.4-2/3.4.5+….+2/98.99.100

   =1/1.2-1/2.3+1/2.3-1/3.4+….+1/98.99-1/99.100

   =1/1.2-1/99.100

   =4950/9900-1/9900

   =4949/9900

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