Question

[Nguyễn Thị Quỳnh Trang]

1+1/3+1/6+1/10+...+2/x.(x+1)=1/2015/2017 (hỗn số)

Answer 1

[soyeon_Tiểu bàng giải]

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=1\frac{2015}{2017}\)

\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{4032}{2017}\)

\(2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{4032}{2017}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{4032}{2017}:2\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4032}{2017}.\frac{1}{2}\)

\(1-\frac{1}{x+1}=\frac{2016}{2017}\)

\(\frac{x}{x+1}=\frac{2016}{2017}\)

=> \(x=2016\)

Answer 2

[Nguyễn Thị Quỳnh Trang]

thanks