Question

cho A=\(\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+7+...+2017}\)

Answer 1

\(A=\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+5+...+2017}\)

\(\Rightarrow A=\frac{1}{\frac{\left(3+1\right).\left[\left(3-1\right):2+1\right]}{2}}+\frac{1}{\frac{\left(5+1\right).\left[\left(5-1\right):2+1\right]}{2}}+...+\frac{1}{\frac{\left(2017+1\right).\left[\left(2017-1\right):2+1\right]}{2}}\)

\(\Rightarrow A=\frac{1}{\frac{4.2}{2}}+\frac{1}{\frac{6.3}{2}}+...+\frac{1}{\frac{2018.1009}{2}}\)

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{1009^2}\)

Answer 2

à còn so sánh A với \(\frac{3}{4}\)nữa