Question

CMR:Với mọi n thuộc N*,ta có:

1/2*5+1/5*8+.....+1/(3n-1)*(3n+2)=n/2*(3n+3)

Answer 1

\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}+\frac{1}{3n+2}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3n+2}\right)=\frac{1}{3}\left(\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right)=\frac{1}{3}\cdot\frac{3n}{2\left(3n+2\right)}=\frac{n}{2\left(3n+2\right)}\)

P/s: pải c/m 1/2*5+1/5*8+.....+1/(3n-1)*(3n+2)=n/2*(3n+2) chứ