Question

Giúp :3

Chứng minh rằng với mọi n \(\varepsilon\) N, ta luôn có:

\(\frac{1}{1.6}\)+ \(\frac{1}{6.11}\) +\(\frac{1}{11.16}\) +...+ \(\frac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\frac{n+1}{5n+6}\) 

B nào nhanh và đúng nhất t tick cho :33

Answer 1

Đặt A = \(\frac{1}{1.6}+\frac{1}{6.11}+..+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)

 5A = \(\frac{5}{1.6}+\frac{5}{6.11}+..+\frac{5}{\left(5n+1\right)\left(5n+6\right)}\)

       = \(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+..+\frac{1}{5n+1}-\frac{1}{5n+6}\)

        = \(\frac{1}{1}-\frac{1}{5n+6}=\frac{5n+6-1}{5n+6}=\frac{5n+5}{5n+6}=\frac{5\left(n+1\right)}{5n+6}\)

=> A  = \(=\frac{5\left(n+1\right)}{5n+6}:5=\frac{5\left(n+1\right)}{5n+6}\cdot\frac{1}{5}=\frac{n+1}{5n+6}\)

VẬy VT = VP ĐT Đ CM