Đặt \(A=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2011}\)
Ta có công thức:
\(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\Rightarrow\frac{1}{1+2+...+n}=\frac{2}{n.\left(n+1\right)}\)
Do đó:
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2011.2012}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2011.2012}\right)\)
\(A=2.\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2012-2011}{2011.2012}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\right)\)
\(A=2.\left(1-\frac{1}{2012}\right)\)
\(A=2.\frac{2011}{2012}\)
\(A=\frac{2011}{1006}.\)
Chúc bạn học tốt!
