câu này quen quen

là s hả bạn?

\(\frac{3}{9.14}+\frac{3}{14.19}+...+\frac{3}{\left(5n-1\right)\left(5n+4\right)}\)

\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{5n-1}-\frac{1}{5n+4}\right)\)

\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{5n+4}\right)\)

\(=\frac{3}{5}.\frac{5n-5}{45n+36}=\frac{n-1}{45n+36}\)

Ta có\(\frac{3}{9.14}+\frac{3}{14.19}+...+\frac{3}{\left(5n-1\right)\left(5n+4\right)}=\frac{3}{5}\left(\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{\left(5n-1\right)\left(5n+4\right)}\right)\)

\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{5n-1}-\frac{1}{5n+4}\right)=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{5n+4}\right)=\frac{1}{15}-\frac{3}{25n+20}\)(1)

kết hợp điều kiện ta có \(\frac{3}{25n+20}\ge\frac{3}{25.2+20}=\frac{3}{70}>0\)

=> \(\frac{3}{9.14}+\frac{3}{14.19}+...+\frac{3}{\left(5n-1\right)\left(5n+4\right)}< \frac{1}{15}\)(đpcm)

Đặt \(A=\frac{3}{9.14}+\frac{3}{14.19}+.......+\frac{3}{\left(5n-1\right)\left(5n+4\right)}\)

\(5A=\frac{15}{9.14}+\frac{15}{14.19}+.....+\frac{15}{\left(5n-1\right)\left(5n+4\right)}\)

\(5A=3.\left(\frac{5}{9.14}+\frac{5}{14.19}+......+\frac{5}{\left(5n-1\right)\left(5n+4\right)}\right)\)

\(5A=3.\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+.....+\frac{1}{5n-1}-\frac{1}{5n+4}\right)\)

\(5A=3.\left(\frac{1}{9}-\frac{1}{5n+4}\right)\)

\(5A=\frac{1}{3}-\frac{1}{5n+4}\)

=> \(5A<\frac{1}{3}\) 

=> \(A<\frac{1}{3}:5\)

hay \(A<\frac{1}{15}\) \(\left(đpcm\right)\)

Nhớ nhé bạn

nhớ bạn

tích  nhé