Câu hỏi của George H. Dalton

Question

Tìm x biết: $\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{4020}{2011}$

Thiên Hàn · Accepted Answer

$\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4020}{2011}$

$\Rightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4020}{2011}$

$\Rightarrow\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4020}{2011}$

$\Rightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{4020}{2011}$

$\Rightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{4020}{2011}:2$

$\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2010}{2011}$

$\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2010}{2011}$

$\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2010}{2011}$

$\Rightarrow\dfrac{1}{x+1}=-\dfrac{2009}{4022}$

$\Rightarrow4022=-2009\left(x+1\right)$

$\Rightarrow4022=-2009x-2009$

$\Rightarrow2009x=-2009-4022$

$\Rightarrow2009x=-6031$

$\Rightarrow x=-\dfrac{6031}{2009}$Câu trả lời: $\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4020}{2011}$