Xét mẫu:

\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+2012}\)

= \(1+\frac{1}{3}+\frac{1}{6}+....+\frac{1}{2025078}\)

= \(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)

= \(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2012}-\frac{1}{2013}\right)\)

= \(1+2.\left(\frac{1}{2013}\right)\)

= \(\frac{4024}{2013}\)

=> E =  \(\frac{2.2012}{\frac{4024}{2013}}\)

=> E = \(4024.\frac{2013}{4024}\)

=> E = 2013

Tính biểu thức D

\(D=\frac{2.2012}{1+\frac{2}{2.\left(1+2\right)}+\frac{2}{2\left(1+2+3\right)}+\frac{2}{2\left(1+2+3+4\right)}+...+\frac{2}{2\left(1+2+..+2012\right)}}\)

\(=\frac{2.2012}{1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{4050156}}\)

\(=\frac{2.2012}{1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{4050156}\right)}\)

\(=\frac{2.2012}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2012.2013}\right)}\)

\(=\frac{2.2012}{1+2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)

\(=\frac{2.2012}{1+2.\left(\frac{1}{2}-\frac{1}{2013}\right)}\)

\(=\frac{2.2012}{1+\frac{2.2011}{2.2013}}\)

\(=\frac{2.2012}{1+\frac{2011}{2013}}\)

\(=\frac{4024}{\frac{4024}{2013}}\)

\(=2013\)

Vậy D=2013

\(D=\frac{2\cdot2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2012}}\)

\(D=\frac{2.2012}{1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2025078}}\)

\(D=\frac{4024}{1+2\cdot\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{4050156}\right)}\)

\(D=\frac{4024}{2\cdot\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2012\cdot2013}\right)}\)

\(D=\frac{4024}{1+2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)

\(D=\frac{4024}{1+2\cdot\left(\frac{1}{2}-\frac{1}{2013}\right)}\)

\(D=\frac{4024}{1+2.\left(\frac{2013-2}{4026}\right)}\)

\(D=\frac{4024}{1+2\cdot\frac{2011}{4026}}\)

\(D=\frac{4024}{1+\frac{2011}{2013}}\)

\(D=\frac{4024}{\frac{4024}{2013}}\)

\(D=\frac{1}{2013}\)

1=1*2/2

1+2=2*3/2

1+2+3=3*4/2

...

1+2+3+...+2012=2012*2013/2

Thay vào là ra.

Đặt A là tên biểu thức

Xét mẫu số, ta có: \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}\)

\(=1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}\)

\(=\frac{2}{2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}\)\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\left(1-\frac{1}{2013}\right)=2\cdot\frac{2012}{2013}\)

\(\Rightarrow A=\frac{2.2012}{2\cdot\frac{2012}{2013}}=\frac{2012.2013}{2012}=2013\)

\(\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}}\)

\(=\frac{2.2012}{1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}}\)

\(=\frac{2.2012}{\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}}\)

\(=\frac{2.2012}{2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)

\(=\frac{2.2012}{2.\left(1-\frac{1}{2013}\right)}=\frac{2.2012}{2.\frac{2012}{2013}}=\frac{2012}{\frac{2012}{2013}}=\frac{2012.2013}{2012}=2013\)

Co quy luat nay ne em: 1+2=3=2.3:2; 1+2+3=6=3.4:2;...;1+2+3+...+2012=2012.2013:2

Suy ra ta co:

Mau so cua D=1 + 1/(2.3:2)  +  1/(3.4:2)   +   1/(4.5:2)   +   ....   +   1/(2012.2013:2)

                    =1  +  2/2.3  +  2/3.4   +   2/4.5   +  ....  +   2/2012.2013

                    = 2.[1/2  +  1/2.3  +  1/3.4  +  1/4.5  +  .... +  1/2012.2013]

                    =2.[1/1.2   +  1/2.3   +   1/3.4   +  1/4.5   +  .....   +  1/2012.2013]

                    =2.[1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +....+1/2012 - 1/2013

                    =2[1 - 1/2013]

                    =2.2012/2013

Vay D= 2.2012 / (2.2012:2013)=2013

phần mẫu số có

\(1+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2012\right).2012:2}\)

\(1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{1}{2012.2013}\)

gọi tổng trên là S. lấy S : 2 có

\(S:2=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\)

\(S:2=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\)

\(S:2=\frac{1}{2}+\frac{1}{2}-\frac{1}{2013}\)

Mẫu số = \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}\)

\(=1+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2012\right).2012:2}\)

\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)

\(=2.\left(1-\frac{1}{2013}\right)=\frac{2.2012}{2013}\)

Phân số đề bài cho = \(\frac{2.2012}{\frac{2.2012}{2013}}=2013\)