Question

Tính A = \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.....+\frac{1}{1+2+3+.....+2014}\)

Help me ~~~~

Answer 1

\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{2014}-\frac{1}{2015}\)

\(=\frac{1}{2}-\frac{1}{2015}=\frac{2013}{4030}=>A=\frac{2013}{4030}:2=\frac{2013}{2015}\)

tick nhe

Answer 2

Hình như đây là bài lớp 6 mà