Câu hỏi của Nguyễn Thị Hải Yến

Question

Tính: $B=\dfrac{1}{2010.2009}-\dfrac{1}{2009.2008}-\dfrac{1}{2008.2007}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}$

Đạt Trần Tiến · Accepted Answer

Ta có B= $\frac{1}{2009.2010}-(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}+\frac{1}{2008.2009}) $

=$\frac{1}{2009.}-\frac{1}{2010} -(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2007}-\frac{1}{2008} +\frac{1}{2008}-\frac{1}{2009}) $

=$\frac{1}{2009}-\frac{1}{2010}-(1-\frac{1}{2009} )$

=$\frac{2}{2009}-1 -\frac{1}{2010} $

Câu trả lời:

Ta có B= $\frac{1}{2009.2010}-(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}+\frac{1}{2008.2009}) $

=$\frac{1}{2009.}-\frac{1}{2010} -(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2007}-\frac{1}{2008} +\frac{1}{2008}-\frac{1}{2009}) $

=$\frac{1}{2009}-\frac{1}{2010}-(1-\frac{1}{2009} )$

=$\frac{2}{2009}-1 -\frac{1}{2010} $