Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)

Ta có: \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=-\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}-\dfrac{1}{4\cdot5}+...-\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

\(\Leftrightarrow2A=-\dfrac{1}{2}+\dfrac{1}{99\cdot100}\)

\(\Leftrightarrow2A=\dfrac{-1}{2}+\dfrac{1}{9900}\)

\(\Leftrightarrow2A=\dfrac{-4950}{9900}+\dfrac{1}{9900}=\dfrac{-4949}{9900}\)

hay \(A=\dfrac{-4949}{19800}\)

Đặt \(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)

\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{30\times31\times32}\)

\(=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+\left(\frac{1}{3\times4}-\frac{1}{4\times5}\right)+...+\left(\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)

\(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\)

\(=\frac{1}{1\times2}-\frac{1}{31\times32}\)

\(=\frac{1}{2}-\frac{1}{992}\)

\(=\frac{495}{992}\)

\(\Rightarrow A=\frac{495}{992}\div2=\frac{495}{1984}\)

\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)

\(=\frac{1}{2}\times\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{31\times32}\right)\)

\(=\frac{1}{2}\times\frac{990}{1984}\)

\(=\frac{990}{3968}=\frac{495}{1984}\)

Tham khảo bài mik tại link này nhé :

https://olm.vn/hoi-dap/detail/197648362685.html

S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2013.2014.2015}\)

S = \(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+....+\frac{2015-2013}{2013.2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+...+\frac{2015}{2013.2014.2015}-\frac{2013}{2013.2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2014.2015}\right)\)

S = \(\frac{1}{2}.\frac{2029104}{4058210}\)

S = \(\frac{1014552}{4058210}\)

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

4A=1.2.3.4+2.3.4(5-1)+3.4.5(6-2)+.....+98.99.100(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+....+98.99.100.101-97.98.99.100

4A=98.99.100.101

A=(98.99.100.101):4=24497550

Cứ một dãy số  thì có 2 thừa số bị gạch nên cuối cùng chỉ còn 1x100

[ 98.99.100.101 - 0.1.2.3]: 4=24497550

Đặt A = 1.2.3 + 2.3.4 + 3.4.5 +...+ (n - 1)n(n + 1)

4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...+ (n - 1)n(n + 1).4

4A = 1.2.3.(4 - 0) + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) +....+ (n - 1)n(n + 1).[(n + 2) - (n - 2)]

4A = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 +...+ (n - 1)n(n + 1)(n + 2) - (n - 2)(n - 1)n(n + 1)

4A = [1.2.3.4 + 2.3.4.5 + 3.4.5.6 +....+ (n - 1)n(n + 1)(n + 2)] - [0.1.2.3 + 1.2.3.4 + 2.3.4.5 + (n - 2)(n - 1)n(n + 1)]

4A = (n - 1)n(n + 1)(n + 2) - 0.1.2.3

4A = (n - 1)n(n + 1)(n + 2)

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

Đặt C = \(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+...+\frac{1}{30\times31\times32}\)

\(2C=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+...+\frac{2}{30\times31\times32}\)

        \(=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+...+\left(\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)

          \(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\)

            \(=\frac{1}{1\times2}-\frac{1}{31\times32}\)

              \(=\frac{1}{2}-\frac{1}{992}=\frac{495}{992}\)

\(\Rightarrow C=\frac{495}{992}\div2=\frac{495}{1984}\)

Vậy ...

\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+.....+\frac{1}{30\times31\times32}\)

\(=\frac{1}{2}\times\left(\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+.....+\frac{2}{30\times31\times32}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{30.31}-\frac{1}{31.32}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{31.32}\right)=\frac{1}{2}.\frac{990}{1984}=\frac{990}{3968}\)

=1+1/32

=33/32