A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)

= \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)

= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)

= \(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1482}\right)\)

= \(\frac{1}{2}.\frac{390}{781}=\frac{195}{781}\)

=195/781

\(n\left(n+1\right)\left(n+2\right)=\frac{1}{4}n\left(n+1\right)\left(n+2\right).4=\frac{1}{4}n\left(n+1\right)\left(n+2\right)\left[\left(n+3\right)-\left(n-1\right)\right]\)

\(=-\frac{1}{4}\left(n-1\right)n\left(n+1\right)\left(n+2\right)+\frac{1}{4}n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)

\(4S=-0.1.2.3+1.2.3.4-1.2.3.4+2.3.4.5-....-\left(k-1\right)k\left(k+1\right)\left(k+2\right)+k\left(k+2\right)\left(k+2\right)\left(k+3\right)\)

\(=k\left(k+1\right)\left(k+2\right)\left(k+3\right)\)

\(4S+1=\left(k^2+3k\right)\left(k^2+3k+2\right)+1=\left(k^2+3k\right)^2+2.\left(k^2+3k\right)+1\)

\(=\left(k^2+3k+1\right)^2\)

ta có:
4s=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+.........+k(k+1)(k+2)((k+3)-(k-1))
4s=1.2.3.4-1.2.3.0+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+........+k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)
4s=k(k+1)(k+2)(k+3)
ta biết rằng tích 4 số tự nhiên liên tiếp khi cộng thêm 1 luôn là 1 số chính phương
=>4s+1 là 1 số chính phương

Ta có:

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{998.999.1000}\)

\(\Rightarrow\frac{1}{2}A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{998.999.1000}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{998.999}-\frac{1}{999.1000}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{999.1000}=\frac{499499}{999000}\Leftrightarrow A=\frac{499499}{1998000}\)

\(B=\frac{1}{1.2.3.4.5}+\frac{1}{2.3.4.5.6}+\frac{1}{3.4.5.6.7}+\frac{1}{996.997.998.999.1000}\)

\(\Rightarrow\frac{1}{4}B=\frac{4}{1.2.3.4.5}+\frac{4}{2.3.4.5.6}+\frac{4}{3.4.5.6.7}+....+\frac{4}{996.997.998.999.1000}\)

\(\Rightarrow\frac{1}{4}B=\frac{1}{1.2.3.4}-\frac{1}{2.3.4.5}+\frac{1}{2.3.4.5}-\frac{1}{3.4.5.6}+\frac{1}{3.4.5.6}-\frac{1}{4.5.6.7}+...+\frac{1}{996.997.998.999}-\frac{1}{997.998.999.1000}\)

\(\Rightarrow\frac{1}{4}B=\frac{1}{1.2.3.4}-\frac{1}{997.998.999.1000}=\frac{41417124749}{994010994000}\Leftrightarrow B=\frac{41417124749}{3976043976000}\)

ta có:
4s=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+.........+k(k+1)(k+2)((k+3)-(k-1))
4s=1.2.3.4-1.2.3.0+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+........+k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)
4s=k(k+1)(k+2)(k+3)
ta biết rằng tích 4 số tự nhiên liên tiếp khi cộng thêm 1 luôn là 1 số chính phương
=>4s+1 là 1 số chính phương

Bạn xem lại đề, là cộng mới đúng chứ ???

Mình làm được rồi này :

\(B=\frac{1}{1.2.3}-\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{97.98.99}\right)\)

    \(=\frac{1}{6}-\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{97.98}-\frac{1}{98.99}\right)\)

    \(=\frac{1}{6}-\left(\frac{1}{2.3}-\frac{1}{98.99}\right)\)

     \(=\frac{1}{6}-\frac{1}{6}+\frac{1}{9702}\)

     \(=\frac{1}{9702}\)