Ta có: \(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{\left(5n+1\right).\left(5n+6\right)}\)

=\(\dfrac{1}{5}.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)

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

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

=\(\dfrac{1}{5}.\dfrac{5n+5}{5n+6}\)

=\(\dfrac{1}{5}.\dfrac{5.\left(n+1\right)}{5n+6}\)

=\(\dfrac{n+1}{5n+6}\left(ĐPCM\right)\)

Ta có :A= \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

2A= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)

2A-A=\(1-\dfrac{1}{2^9}\)

A=1-\(\dfrac{1}{2^9}\)

Mình biết bao nhiêu thì mình giải bấy nhiêu chúc bạn học tốt nhé