Question

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

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

Answer 1

\(\dfrac{1}{1\cdot6}+\dfrac{1}{6\cdot11}+\dfrac{1}{11\cdot16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)

\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)

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

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

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

Answer 2

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

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

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

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

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

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

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