Ta thấy: \(1+2+3+...+n=\dfrac{n\cdot\left(n+1\right)}{2}\left(n\in N^{\circledast}\right)\)
\(\Rightarrow2\cdot\left(1+2+3+...+n\right)=n\cdot\left(n+1\right)\)
\(H=2012-\left(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+99}\right)\\ =2012-\left(1+\dfrac{2}{2\cdot\left(1+2\right)}+\dfrac{2}{2\cdot\left(1+2+3\right)}+...+\dfrac{2}{2\cdot\left(1+2+3+...+99\right)}\right)\\ =2012-\left(\dfrac{2}{2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\right)\\ =2012-2\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\\ =2012-2\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2012-2\cdot\left(1-\dfrac{1}{100}\right)\\ =2012-2\cdot\dfrac{99}{100}\\ =2012-\dfrac{99}{50}\\ =2010\dfrac{1}{50}\)
