Question

Biểu thức A = \(\frac{1}{1+2}+\frac{1}{1+2+3}+......+\frac{1}{1+2+3+4+...2014}\)   có giá trị bằng

Answer 1

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}\)

\(A=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2014\right).2014:2}\)

\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2014.2015}\)

\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{2015}\right)\)

\(A=2.\frac{1}{2}-2.\frac{1}{2015}\)

\(A=1-\frac{2}{2015}=\frac{2013}{2015}\)