Question

Tìm x:

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\)\(\frac{1}{x\left(x+1\right)}\)=\(\frac{2015}{2016}\)

Answer 1

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2015}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(1-\frac{1}{x+1}=1-\frac{2015}{2016}\)

\(\frac{1}{x+1}=\frac{1}{2016}\)

\(x=2016-1\)

\(\Rightarrow x=2015\)