Question

[cao xuan phong]

Cho \(b=\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+...+101}\)

Chứng minh \(b< \frac{3}{4}\)

Answer 1

[nguyen thi hien]

\(B=\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+...+101}\)

\(B=\frac{1}{4}+\frac{1}{9}+...+\frac{1}{51}\)

\(B=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+...+\frac{1}{3\cdot17}\)

\(B=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{3}-\frac{1}{17}\)

\(B=\frac{1}{2}-\frac{1}{17}\)

\(B=\frac{15}{34}\)

TU DO \(=>\frac{15}{34}< \frac{3}{4}\)HOAC \(B< \frac{3}{4}\)

 CHUC BAN HOC TOT :)) 

Answer 2

[Đông Phương Lạc]

Ta có: \(1+3=\frac{\left(1+3\right).\left[\left(3-1\right):2+1\right]}{2}=\frac{4.2}{2}=2.2\)

\(1+3+5=\frac{\left(1+5\right).\left[\left(5-1\right):2+1\right]}{2}=\frac{6.3}{2}=3.3\)

                  \(.................\)

\(1+3+5+...+101=\frac{\left(1+101\right).\left[\left(101-1\right):2+1\right]}{2}=\frac{102.5}{2}=51.51\)

\(\Rightarrow B=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{51.51}\)

\(\Rightarrow B< \frac{1}{2.2}+\frac{1}{2.3}+...+\frac{1}{50.51}=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{50}-\frac{1}{51}\)

\(\Rightarrow B< \left(\frac{1}{4}+\frac{1}{2}\right)-\frac{1}{51}\)

\(\Rightarrow B< \frac{3}{4}-\frac{1}{51}< \frac{3}{4}\)

\(\Rightarrow B>\frac{3}{4}\left(đpcm\right)\)