Question

Chứng minh rằng với mọi \(n\in N\); \(n\ge2\) ta có :

\(\dfrac{3}{9.14}+\dfrac{3}{14.19}+\dfrac{3}{19.24}+..........+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}< \dfrac{1}{15}\)

Help me!!!!!!!!!!!!!!!!!!

Answer 1

thoi lay cai nich nguyen thanh hang dang len di

lay nich nay de do nhuc mat chu ji

Answer 2

help me!!!!!!!!!!!!!!!!!!!! lam ji

Answer 3

cac ban thay anh nen cua toi dep ko

Answer 4

Ra lớp thầy chữa cho

Answer 5

\(VT=\dfrac{3}{9\cdot14}+\dfrac{3}{14\cdot19}+...+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}\)

\(=\dfrac{3}{5}\left(\dfrac{5}{9\cdot14}+\dfrac{5}{14\cdot19}+...+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}\right)\)

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

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

\(=\dfrac{3}{5}\cdot\dfrac{5\left(n-1\right)}{9\left(5n+4\right)}\)\(=\dfrac{n-1}{3\left(5n+4\right)}< \dfrac{1}{15}=VP\forall n\in N;n\ge2\)