D = 1.2 + 2.3+ 3.4 +...+ 99.100

=>3D=1.2.3+2.3.3+3.4.3+...+99.100.3

=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+....+99.100.(101-98)

=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

=99.100.101-0.1.2

=99.100.101

=999900

=>D=999900:3=333300

Dn = 1.2 + 2.3 + 3.4 +...+ n (n +1)

=>3Dn=1.2.3+2.3.3+3.4.3+...+n(n+1).3

=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+n.(n+1).[(n+2)-(n-1)]

=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-2.3.4+....+n(n+1)(n+2)-(n-1)n(n+1)

=n.(n+1).(n+2)-0.1.2

=n.(n+1)(n+2)

=>Dn=n.(n+1)(n+2):3

 =>điều cần chứng minh

Đặt \(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2.}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(\Rightarrow\frac{S}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(=\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\frac{1}{x+1}=\frac{1}{2001}\)

\(\Rightarrow\)x+1=2001

x=2000

Vậy x=2000.

Cách lớp 7 nà:)

\(\frac{1}{n.\left(n+1\right)^2}=\frac{1}{n.\left(n+1\right).\left(n+1\right)}< \frac{1}{n.n\left(n+1\right)}< \frac{1}{\left(n-1\right)n\left(n+1\right)}\) (n>=2_

\(\text{Suy ra }VT< \frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)

Mặt khác ta có công thức \(\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]}{2}\) (n>= 2)

Suy ra \(VT< \frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{n\left(n+1\right)}\right)< \frac{1}{2}.\frac{1}{2}=\frac{1}{4}\left(\text{do }\frac{1}{n\left(n+1\right)}>0\right)\)

Vậy ta có đpcm

Gắt chưa??? :>> Dương Bá Gia Bảo

mk làm đc rồi nha

Ta có :

\(A=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+...+\frac{\left(n-1\right)n-1}{n!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{\left(n-1\right)n}{n!}-\frac{1}{n!}\)

\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4}!+\frac{1}{3!}-\frac{1}{5!}+\frac{1}{4!}-...+\frac{1}{\left(n-2\right)!}-\frac{1}{n!}\)

\(=2-\frac{1}{n!}< 2\)

Vậy ...

3S = 1.2.3+2.3.3+3.4.3+.....+n.(n+1).3

= 1.2.3+2.3.(4-1)+3.4.(5-2)+.....+n.(n+1).[(n+2)-(n-1)]

= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....+n.(n+1).(n+2)-(n-1).n.(n+1)

= n.(n+1).(n+2)

=> 3S +n.(n+1).(n^2-2) = n.(n+1).(n+2)+n.(n+1).(n^2-2)

= n.(n+1).(n+2+n^2-2) = n.(n+1).(n^2+n)

= n.(n+1)+n.(n+1) = n^2.(n+1)^2 = [(n.(n+1)]^2 là 1 số chính phương

k mk nha